Let the height of the tower = h m, and distance of point P from the foot of the tower = d m.

Let AB = tower (height = h), BC = pole (height = 6 m), P = point on ground.

From the angle of depression of P from the top of the tower (45°):

$$\tan 45° = \frac{h}{d} \implies 1 = \frac{h}{d} \implies d = h \quad \text{...(1)}$$

From the angle of elevation of the top of the pole from P (60°):

$$\tan 60° = \frac{h + 6}{d} \implies \sqrt{3} = \frac{h + 6}{d} \quad \text{...(2)}$$

Substituting (1) into (2):

$$\sqrt{3} = \frac{h + 6}{h}$$

$$\sqrt{3}\, h = h + 6$$

$$h(\sqrt{3} - 1) = 6$$

$$h = \frac{6}{\sqrt{3} - 1} = \frac{6(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{6(\sqrt{3} + 1)}{2} = 3(\sqrt{3} + 1)$$

$$h = 3(1.73 + 1) = 3 \times 2.73 = \mathbf{8.19 \text{ m}}$$

Distance of P from foot of tower:

$$d = h = \mathbf{8.19 \text{ m}}$$

∴ Height of tower = 8.19 m and distance of point P from the foot of the tower = 8.19 m.

Source: Chapter 9, Section 9.1 – Heights and Distances

---

• The angle of depression from the top of the tower to P equals the angle of elevation from P to the top of the tower (alternate angles), giving tan 45° = h/d → d = h.
• The angle of elevation to the top of the pole uses the total height (h + 6), giving tan 60° = (h+6)/d.
• Rationalise the denominator when solving for h. Examiners award marks for the diagram, correct equations, solving, and substituting √3 = 1.73 at the end.

(i) From window W, angle of depression of foot B is 30°.

$$\tan 30° = \frac{d}{54} \Rightarrow \frac{1}{\sqrt{3}} = \frac{d}{54}$$

So the relation is: $d = \frac{54}{\sqrt{3}}$ i.e., $y = d = \frac{54}{\sqrt{3}} = 18\sqrt{3}$ m

(ii) From window W, angle of elevation of top A is 45°.

$$\tan 45° = \frac{h}{54} \Rightarrow 1 = \frac{h}{54} \Rightarrow h = 54 \text{ m}$$

(iii) Height of water tank AB = $h + d = 54 + 18\sqrt{3}$ m

$$= 18(3 + \sqrt{3}) = 18\sqrt{3}(\sqrt{3}+1) \text{ m}$$

OR

$x = 54$ m (horizontal distance, as $\tan 45° = h/x \Rightarrow x = h = 54$ m)

Height of window above ground $= d = 18\sqrt{3}$ m $\approx 31.18$ m

Source: Heights and Distances, Application of Trigonometry

---

• The horizontal distance (54 m) is used with both angles to set up tan equations.
• $h$ is the vertical height from window level up to top of tank; $d$ is from ground to window (same as depression side).
• Total tank height = $h + d$ since B is at ground and A is at top.
• Examiners expect clear trigonometric ratios, substitution, and final simplified value.

(A) 10 m

In right △ABC, AB = 5√3 m (base), angle of elevation = 60°.
cos 60° = AB/AC → 1/2 = 5√3/AC → AC = 10 m.

The wire is the hypotenuse (AC). Use cos 60° = adjacent/hypotenuse = AB/AC. Since cos 60° = 1/2, AC = 2 × 5√3 is wrong — note AB here is the horizontal distance, so AC = AB/cos60° = 5√3 ÷ (1/2) = 10 m. Always identify which side is the hypotenuse (the wire/rope) and which trig ratio links the given side and required side.

Diagram: Let AB = 8 m be the building, CD be the cable tower. Draw BE ∥ AC. Then BE = AC (horizontal distance) and CE = AB = 8 m.

Step 1: Find horizontal distance (AC)

Angle of depression of foot of tower = 45°
In △ABC (right-angled at C):
$$\tan 45° = \frac{AB}{AC} \Rightarrow 1 = \frac{8}{AC} \Rightarrow AC = 8 \text{ m}$$

So BE = AC = 8 m.

Step 2: Find DE (height above building level)

Angle of elevation of top of tower = 60°
In △BED (right-angled at E):
$$\tan 60° = \frac{DE}{BE} \Rightarrow \sqrt{3} = \frac{DE}{8} \Rightarrow DE = 8\sqrt{3} \text{ m}$$

Step 3: Total height of tower

$$CD = CE + DE = 8 + 8\sqrt{3} = 8(1 + \sqrt{3})$$
$$= 8(1 + 1.732) = 8 \times 2.732 = \textbf{21.856 m}$$

The height of the cable tower is $8(1+\sqrt{3})$ m ≈ 21.856 m.

Source: Chapter 9, Exercise 9.1 (Q.12), Heights and Distances

---

• Draw a horizontal line from the top of the building to create two separate right triangles — this is the key construction step examiners look for.
• Angle of depression (45°) gives the horizontal distance; angle of elevation (60°) gives the extra height above the building.
• Total tower height = height of building + extra height = 8 + 8√3.
• Always substitute √3 = 1.732 and compute the numerical value when the question asks for it.
• The textbook Q.12 uses a 7 m building; this variant uses 8 m — the method is identical.

Let the point on the ground be O, and let the horizontal distance from O to the point directly below both planes be $d$ metres.

For the lower plane (height = 3125 m, angle of elevation = 30°):

$$\tan 30° = \frac{3125}{d} \implies \frac{1}{\sqrt{3}} = \frac{3125}{d} \implies d = 3125\sqrt{3} \text{ m}$$

For the upper plane (height = $h$ m, angle of elevation = 60°):

$$\tan 60° = \frac{h}{d} \implies \sqrt{3} = \frac{h}{3125\sqrt{3}} \implies h = 3125\sqrt{3} \times \sqrt{3} = 9375 \text{ m}$$

Distance between the two planes:

$$= 9375 - 3125 = \boxed{6250 \text{ m}}$$

Source: Chapter 9, Heights and Distances

---

• Both planes fly vertically above the same point, so both share the same horizontal distance $d$ from the observer.
• Use $\tan\theta = \frac{\text{height}}{d}$ for each plane separately.
• Find $d$ from the lower plane's data, then find the upper plane's height using the same $d$.
• The answer is simply the difference in heights. Examiners award marks for the diagram/setup (1), correct use of tan for each plane (1), and the final subtraction (1).

(1) Length of strings CA and CB:

For kite A: angle of elevation = 30°, AD = 50 m

$$\sin 30° = \frac{AD}{CA} \Rightarrow \frac{1}{2} = \frac{50}{CA} \Rightarrow CA = 100 \text{ m}$$

For kite B: angle of elevation = 60°, BE = 60 m

$$\sin 60° = \frac{BE}{CB} \Rightarrow \frac{\sqrt{3}}{2} = \frac{60}{CB} \Rightarrow CB = \frac{120}{\sqrt{3}} = 40\sqrt{3} \text{ m}$$

(2) Distance 'd' between the two kites:

$$CD = \frac{AD}{\tan 30°} = \frac{50}{1/\sqrt{3}} = 50\sqrt{3} \text{ m}$$

$$CE = \frac{BE}{\tan 60°} = \frac{60}{\sqrt{3}} = 20\sqrt{3} \text{ m}$$

Since D, C, E are collinear and A, B are above D and E respectively:

$$AB = d = \sqrt{(AD - BE)^2 + (DC + CE)^2}$$

$$= \sqrt{(50-60)^2 + (50\sqrt{3}+20\sqrt{3})^2} = \sqrt{100 + (70\sqrt{3})^2}$$

$$= \sqrt{100 + 14700} = \sqrt{14800} = 20\sqrt{37} \text{ m}$$

Source: Chapter 9 – Some Applications of Trigonometry

---

• For string length, use sin θ = perpendicular/hypotenuse (hypotenuse = string = line of sight).
• For horizontal distance (DC, CE), use tan θ = perpendicular/base.
• Since D and E are on opposite sides of C, total horizontal separation = DC + CE.
• Use Pythagoras for the slant distance AB, as A and B are at different heights.
• Key values: sin 30° = ½, sin 60° = √3/2, tan 30° = 1/√3, tan 60° = √3.

Let AB = 50 m be the tower. Let C and D be the two cars on opposite sides.

For car C (angle of depression = 30°):

In △ABC, ∠ACB = 30° (alternate angles)

$$\tan 30° = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{50}{BC}$$

$$BC = 50\sqrt{3} \text{ m}$$

For car D (angle of depression = 45°):

In △ABD, ∠ADB = 45°

$$\tan 45° = \frac{AB}{BD} \implies 1 = \frac{50}{BD}$$

$$BD = 50 \text{ m}$$

Distance between the two cars:

$$CD = BC + BD = 50\sqrt{3} + 50 = 50(\sqrt{3} + 1) \text{ m}$$

Source: Chapter 9, Heights and Distances

---

• The angle of depression equals the alternate interior angle at the base (since the horizontal from the top is parallel to the ground). Always state this step — examiners award a mark for it.
• Use tan (opposite/adjacent) since height and base distance are involved.
• Final answer must be left as $50(\sqrt{3}+1)$ m or approximately 136.6 m. Both forms are acceptable; exact form is preferred.
• 3 marks are typically split: 1 mark for correct triangle/setup, 1 mark for BC, 1 mark for BD and final answer.

Let the base of the tower be P and PO = 36 cm.

(i) Length of wire from O to top of Section B:

Let top of Section B = Q. Angle of elevation = 30°.

$$\cos 30° = \frac{PO}{OQ} \Rightarrow \frac{\sqrt{3}}{2} = \frac{36}{OQ} \Rightarrow OQ = \frac{72}{\sqrt{3}} = 24\sqrt{3} \text{ cm}$$

Length of wire = 24√3 cm

(ii) Distance AB:

Height of Section B (PQ):
$$\tan 30° = \frac{PQ}{36} \Rightarrow PQ = \frac{36}{\sqrt{3}} = 12\sqrt{3} \text{ cm}$$

Height of top of Section A (PA):
$$\tan 45° = \frac{PA}{36} \Rightarrow PA = 36 \text{ cm}$$

$$AB = PA - PQ = 36 - 12\sqrt{3} = 12(3 - \sqrt{3}) \text{ cm}$$

(iii) Height of Section A from base:

$$PA = 36 \text{ cm}$$

Height of Section A = 36 cm

Source: Applications of Trigonometry, Chapter 9

---

• Part (i): Use cos ratio (hypotenuse = wire from O to top of B). cos 30° = base/hypotenuse.
• Part (ii): Use tan for both heights, then subtract PQ from PA to get AB. The key step examiners check is the subtraction and simplification 12(3−√3).
• Part (iii): tan 45° = 1, so height = horizontal distance directly. Simple but must be stated clearly.
• Always define your variables and show the trig ratio used — that's where method marks lie.

Let the length of the shadow be $x$ m. The pole of height 18 m is vertical, so:

$$\tan\theta = \frac{\text{Height of pole}}{\text{Length of shadow}} = \frac{18}{x}$$

Given $\tan\theta = \dfrac{6}{7}$:

$$\frac{6}{7} = \frac{18}{x}$$

$$x = \frac{18 \times 7}{6} = 21 \text{ m}$$

∴ The length of the shadow is 21 m.

Source: Some Applications of Trigonometry, Section 9.1

---

• Set up a right triangle: pole = opposite side, shadow = adjacent side, so $\tan\theta = \frac{\text{height}}{\text{shadow}}$.
• Substitute the given values and solve for $x$ by cross-multiplication.
• Examiners expect a clear diagram description (or equation setup) and the final answer with units. Two marks: one for correct equation setup, one for correct answer.

Given: Balloon of radius $r$ subtends angle 60° at observer's eye O. Angle of elevation of centre C is 45°.

To prove: Height of centre = $\sqrt{2}\,r$

Construction/Setup:

Let O be the eye of the observer and C be the centre of the balloon. Draw tangents OA and OB to the balloon from O.

Since the balloon subtends an angle of 60° at O:
$$\angle AOB = 60° \implies \angle AOC = 30°$$

In right $\triangle OAC$ (OA is tangent, so $\angle OAC = 90°$):
$$\sin(\angle AOC) = \frac{AC}{OC}$$
$$\sin 30° = \frac{r}{OC}$$
$$\frac{1}{2} = \frac{r}{OC} \implies OC = 2r$$

Finding height of centre C:

Let M be the point on the ground directly below O (observer's level), and let CM be the height of C above observer's level (taken as horizontal reference). The angle of elevation of C is 45°, so:

$$\sin 45° = \frac{h}{OC}$$

where $h$ = height of centre C above the observer's eye.

$$\frac{1}{\sqrt{2}} = \frac{h}{2r}$$
$$h = \frac{2r}{\sqrt{2}} = \sqrt{2}\,r$$

Hence, the height of the centre of the balloon is $\sqrt{2}$ times its radius. $\blacksquare$

Source: Chapter 9 — Some Applications of Trigonometry, Heights and Distances

---

• The key step is recognising that $OA \perp AC$ (radius to tangent point is perpendicular), letting you use $\sin 30° = r/OC$ to get $OC = 2r$.
• Then use $\sin 45° = h/OC$ (not $\tan$, because $OC$ is a slant distance from observer to centre, not a horizontal base).
• Examiners expect a clear diagram description, correct triangle identification, and all trigonometric steps shown explicitly.
• Avoid confusing $\tan$ (used when base is known) with $\sin$ (used when hypotenuse is known); here $OC$ is the hypotenuse of the elevation triangle.

Let tower AB = 24 m (first tower), tower CD = h m (second tower), and distance between them = BC = d m.

Step 1: Find distance between towers (d)

From foot of tower CD (point C), angle of elevation of top of AB = 60°.

In △ABC:
$$\tan 60° = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{24}{d} \Rightarrow d = \frac{24}{\sqrt{3}} = 8\sqrt{3} \text{ m}$$

Step 2: Find height of second tower (h)

From foot of tower AB (point B), angle of elevation of top of CD = 30°.

In △BCD:
$$\tan 30° = \frac{CD}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{8\sqrt{3}} \Rightarrow h = \frac{8\sqrt{3}}{\sqrt{3}} = 8 \text{ m}$$

Step 3: Find length of wire (AC) joining tops of both towers

$$AC = \sqrt{BC^2 + (AB - CD)^2} = \sqrt{(8\sqrt{3})^2 + (24-8)^2}$$
$$= \sqrt{192 + 256} = \sqrt{448} = 4\sqrt{28} = 8\sqrt{7} \text{ m}$$

Results: Distance between towers = $8\sqrt{3}$ m, height of second tower = 8 m, length of wire = $8\sqrt{7}$ m.

Source: Chapter 9, Section 9.1 Heights and Distances

---

• Draw a clear diagram and label both towers before solving — examiners award a mark for the diagram.
• Use tan ratio since you know the opposite (height) and adjacent (distance) sides.
• For the wire length, apply Pythagoras on the right triangle formed by the horizontal distance and the difference in heights (since tops are at different levels).
• Keep exact surd form unless told to approximate; $\sqrt{448} = \sqrt{64 \times 7} = 8\sqrt{7}$.

(i) Diagram – Situation I:

```

• (Top of Statue)

|
| h
|
|___________
| Base | A (observer on ground)
|___________|←——— 80√3 m ———→
∠60°
```
A right triangle with observer A on ground, horizontal distance = 80√3 m, angle of elevation = 60°.

---

(ii) Diagram – Situation II:

```

• (Top, 240 m)

|
|
|
_____|
| |←——— d ———→ B (40 m above ground)
|240m| ∠30°
|___|
```
Observer B is 40 m above ground; angle of elevation to top = 30°; total height = 240 m.

---

(iii) Calculation using Situation – I:

Let total height of statue (including base) = H m.

$$\tan 60° = \frac{H}{80\sqrt{3}}$$

$$\sqrt{3} = \frac{H}{80\sqrt{3}}$$

$$H = 80\sqrt{3} \times \sqrt{3} = 80 \times 3 = 240 \text{ m}$$

Height including base = 240 m

Height excluding base = 240 – 58 = 182 m

Source: Some Applications of Trigonometry, NCERT Class 10 Maths

---

• For (i) and (ii), even a rough labelled sketch earns the mark — label the angle, distance, and height clearly.
• For (iii), examiners award 1 mark for correct use of tan 60° and setting up the equation, and 1 mark for the final answers (240 m and 182 m). Always subtract the given base height (58 m) to find the statue height alone.
• Key formula: $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$

Option A: 30°

Here, height of tree = 10 m, horizontal distance = $10\sqrt{3}$ m. $\tan\theta = \frac{10}{10\sqrt{3}} = \frac{1}{\sqrt{3}}$, so $\theta = 30°$.

Use $\tan(\text{angle of depression}) = \frac{\text{height}}{\text{horizontal distance}}$. Since the peacock looks down, the angle of depression equals the angle whose tangent is $\frac{10}{10\sqrt{3}} = \frac{1}{\sqrt{3}}$, giving 30°. A common mistake is inverting the ratio and getting 60°—always put height in the numerator and base distance in the denominator when using tan for angle of depression from the top.

(i) Let the angle of inclination be θ.

Given: height = 15 ft, horizontal distance = 15 ft

$$\tan\theta = \frac{15}{15} = 1 \Rightarrow \theta = 45°$$

The stairs are inclined at 45°.

---

(ii) Using Pythagoras theorem:

$$\text{Length of stairs} = \sqrt{15^2 + 15^2} = \sqrt{225 + 225} = \sqrt{450} = 15\sqrt{2} \text{ feet}$$

The length of stairs is $15\sqrt{2}$ feet.

---

(iii) Given: length of stairs = 20 ft, angle = 60°

$$\sin 60° = \frac{\text{height}}{20}$$

$$\frac{\sqrt{3}}{2} = \frac{h}{20}$$

$$h = 10\sqrt{3} = 10 × 1.732 = \textbf{17.32 feet}$$

The height of the door sill is 17.32 feet.

Source: Applications of Trigonometry, Chapter 9

---

• Part (i): Since height = base = 15 ft, tan θ = 1, so θ = 45°. This is a standard trigonometric value.
• Part (ii): The stairs form the hypotenuse; use Pythagoras or note that for a 45° right triangle, hypotenuse = side × √2.
• Part (iii): The stairs are the hypotenuse (20 ft). Use sin (opposite/hypotenuse) since height is opposite the given angle. Substitute √3 = 1.732 as instructed. Examiners award 1 mark for correct formula and 1 mark for final answer.

Option C: 40 m

Using tan 45° = 1, the horizontal distance from observer's eyes to chimney = 38.2 m. So the chimney height above eye level = 38.2 m. Total height = 38.2 + 1.8 = 40 m.

This follows the same method as Example 3 in the textbook. The observer's eye level is 1.8 m above the ground. Since tan 45° = 1, the vertical rise from eye level equals the horizontal distance (38.2 m). Add the observer's height (1.8 m) to get the total chimney height: 38.2 + 1.8 = 40 m. Always remember to add the observer's height to the calculated vertical distance to get the height above ground.

(i) Diagram:

```
Q
| \
| \
100m | \ (angle of depression = 30°)
| \
P__________A
```
Q is directly above P (QP = 100 m), A is the ambulance on the highway.

---

(ii) Let A be the position of the ambulance. In △QPA:

$$\tan 30° = \frac{QP}{PA} = \frac{100}{PA}$$

$$PA = \frac{100}{\tan 30°} = 100\sqrt{3} = 100 \times 1.73 = \textbf{173 m}$$

---

(iii) When angle of depression = 45°, let the new position be A′:

$$\tan 45° = \frac{100}{PA'} \Rightarrow PA' = 100 \text{ m}$$

Distance covered by ambulance = PA − PA′ = 173 − 100 = 73 m

Speed = 60 km/h $= 60 \times \dfrac{1000}{3600} = \dfrac{50}{3}$ m/s

$$\text{Time} = \frac{73}{\frac{50}{3}} = \frac{73 \times 3}{50} = \frac{219}{50} = \textbf{4.38 seconds}$$

• (i) A right-angle triangle with vertical side QP = 100 m and horizontal side PA is expected. Even a rough labelled sketch scores the mark.
• (ii) Use tan(angle) = opposite/adjacent. The angle of depression from Q equals the angle of elevation from A (alternate interior angles), so tan 30° = 100/PA directly.
• (iii) Find PA′ using tan 45° = 1, subtract to get distance covered, convert speed to m/s, then apply Time = Distance/Speed. Show every step clearly for full 2 marks.

Distance walked by boy in 3 sec = 2.5 × 3 = 7.5 m from the base of the lamp post.

Let the length of shadow = $x$ m.

Using similar triangles (lamp post and boy):

$$\frac{12}{1.5} = \frac{7.5 + x}{x}$$

$$8x = 7.5 + x$$

$$7x = 7.5$$

$$x = \frac{7.5}{7} = \frac{15}{14} \approx 1.07 \text{ m}$$

∴ Length of shadow after 3 seconds = $\dfrac{15}{14}$ m ≈ 1.07 m

• The key step is setting up similar triangles: the lamp post (12 m) and the boy (1.5 m) are parallel verticals; their tips and the shadow tip are collinear, giving the ratio 12/1.5 = (total distance from post to shadow tip)/(shadow length).
• First find the distance of the boy from the post (speed × time = 7.5 m), then let shadow = $x$ so total = $7.5 + x$.
• Examiners award 1 mark for correct similar-triangle setup and 1 mark for correct final answer.

(d) $15\sqrt{3}$ m

Using $\sin 60° = \dfrac{\text{height}}{\text{hypotenuse}}$: height $= 30 \times \sin 60° = 30 \times \dfrac{\sqrt{3}}{2} = 15\sqrt{3}$ m.

The rope is the hypotenuse (30 m) and makes 60° with the ground. The pole (opposite side) = hypotenuse × sin 60°. Students sometimes confuse sin and tan here — use sin because hypotenuse is known, not the base.

Let the height of the kite from the ground = h m.

For Boy 1 (on ground):
String length = 60 m, angle of elevation = 30°

$$\sin 30° = \frac{h}{60} \implies \frac{1}{2} = \frac{h}{60} \implies h = 30 \text{ m}$$

∴ Height of kite from ground = 30 m

For Boy 2 (on roof of 20 m building):
Height of kite above Boy 2's level = 30 − 20 = 10 m
Angle of elevation = 45°

$$\tan 45° = \frac{10}{d} \implies 1 = \frac{10}{d} \implies d = 10 \text{ m}$$

where d = horizontal distance of kite from the building.

Total distance of Boy 1 from base of building:

Let horizontal distance of kite from Boy 1 = x.

$$\cos 30° = \frac{x}{60} \implies \frac{\sqrt{3}}{2} = \frac{x}{60} \implies x = 30\sqrt{3} \text{ m}$$

Since both boys are on opposite sides of the kite:

$$\text{Distance of Boy 1 from base of building} = x + d = 30\sqrt{3} + 10$$
$$= 30 \times 1.73 + 10 = 51.9 + 10 = \textbf{61.9 m}$$

Height of kite from ground = 30 m; Distance of Boy 1 from base of building = 61.9 m.

Source: Chapter 9, Section 9.1 Heights and Distances

---

• Use sin for Boy 1 (string is hypotenuse, height is opposite side).
• For Boy 2, subtract building height from kite height to get the vertical difference, then use tan 45° = 1.
• Use cos 30° to find the horizontal distance from Boy 1 to the point below the kite.
• Since the boys are on opposite sides, add the two horizontal distances to get the total distance between Boy 1 and the building base.
• Examiners expect a clear diagram (optional but helpful), labelled equations, and substitution of √3 = 1.73 at the final step.

Option B: 30°

Let height of tower = h, shadow length = √3h. Then tan θ = h/(√3h) = 1/√3, so θ = 30°.

The altitude of the Sun equals the angle of elevation from the tip of the shadow to the top of the tower. Using tan θ = height/shadow = h/(√3h) = 1/√3 gives θ = 30°. Remember: tan 30° = 1/√3.

Let AB = height of building, C and D be the two positions of the car (C nearer, D farther). CB = 25 m.

(i) Height of the building:

From △ABC: tan 60° = AB/BC
⟹ AB = 25√3 m

(ii) Distance between the two positions:

From △ABD: tan 30° = AB/BD
⟹ BD = AB/tan 30° = 25√3 × √3 = 75 m

Distance CD = BD − BC = 75 − 25 = 50 m

(iii) Total time to reach foot of building from D:

Speed of car = CD/time = 50/6 = 25/3 m/s

Total distance from D to B = 75 m

Total time = 75 ÷ (25/3) = 75 × 3/25 = 9 seconds

---

• Examiner expects correct tan ratio setup for both angles.
• Height comes directly from the nearer triangle (60°, 25 m).
• For part (iii), find speed from the 50 m in 6 s, then apply it to the full 75 m — many students forget to use the total distance DB, not just DC.

Option C: 60

Using $\tan 45° = \dfrac{\text{height}}{\text{distance}} = \dfrac{h}{60}$, and since $\tan 45° = 1$, we get $h = 60$ m.

Apply $\tan(\text{angle of elevation}) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{h}{60}$. Since $\tan 45° = 1$, height = 60 m directly. No surds involved — a common trap is confusing this with 30° or 60° cases.

Given: PO = 6 m, angle of elevation of B = 30°, angle of elevation of A = 60°

(i) Length of wire OB:

In △OPB: cos 30° = PO/OB

$$OB = \frac{PO}{\cos 30°} = \frac{6}{\frac{\sqrt{3}}{2}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \text{ m}$$

(ii) Length of wire OA:

In △OPA: cos 60° = PO/OA

$$OA = \frac{PO}{\cos 60°} = \frac{6}{\frac{1}{2}} = 12 \text{ m}$$

(iii) Distance AB:

$$PB = PO \times \tan 30° = 6 \times \frac{1}{\sqrt{3}} = 2\sqrt{3} \text{ m}$$

$$PA = PO \times \tan 60° = 6 \times \sqrt{3} = 6\sqrt{3} \text{ m}$$

$$AB = PA - PB = 6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3} \text{ m}$$

Source: Applications of Trigonometry, Chapter 9

---

• For wire length, use cos θ = base/hypotenuse (hypotenuse = wire = OB or OA).
• For heights PB and PA, use tan θ = height/base.
• AB = PA − PB since B is below A on the same vertical tower.
• Examiner awards 1 mark each for (i) and (ii), and in (iii): 1 mark for finding both heights correctly + 1 mark for the subtraction. Show all steps clearly.

(d) 60°

Using $\tan\theta = \dfrac{\text{height}}{\text{distance}} = \dfrac{30}{10\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$, so $\theta = 60°$.

Set up a right triangle: tower height = 30 m (perpendicular), base distance = 10√3 m. Apply tan(angle of elevation) = opposite/adjacent = 30/10√3 = √3, which corresponds to 60°. This is a direct application of Example 1-style problems from Chapter 9.

Diagram: Let AB = h (height of tower), C and D be the initial and later positions of the car, with D closer to the foot B.

Setting up equations:

In △ABD (angle of depression = 60°):
$$\tan 60° = \frac{h}{BD} \Rightarrow \sqrt{3} = \frac{h}{BD} \Rightarrow BD = \frac{h}{\sqrt{3}}$$

In △ABC (angle of depression = 30°):
$$\tan 30° = \frac{h}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{BC} \Rightarrow BC = h\sqrt{3}$$

Distance covered in 10 seconds:
$$CD = BC - BD = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$$

Time to travel remaining distance BD:

Speed $= \dfrac{CD}{10} = \dfrac{2h}{10\sqrt{3}}$

$$\text{Time} = \frac{BD}{\text{speed}} = \frac{\dfrac{h}{\sqrt{3}}}{\dfrac{2h}{10\sqrt{3}}} = \frac{h}{\sqrt{3}} \times \frac{10\sqrt{3}}{2h} = 5 \text{ seconds}$$

The car will reach the foot of the tower in 5 seconds.

Source: Chapter 9 (Some Applications of Trigonometry), Exercise 9.1

---

• Note: The textbook (Q.15) uses 6 seconds, giving answer 3 seconds. This question uses 10 seconds, giving 5 seconds. The method is identical — only the numbers differ.
• Key step: angles of depression from the top equal angles of elevation from the base (alternate angles), so use tan directly in the right triangles.
• The ratio CD : BD = 2 : 1, so time for BD = half the time for CD. Examiners award marks for correct tan equations, correct CD, correct speed ratio, and final answer.

Let the two poles AB = 20 m and CD = 28 m be standing vertically. The wire connects their tops, i.e., A and C.

Difference in heights = 28 − 20 = 8 m = CE (where E is the point on CD such that AE is horizontal).

In right △AEC, the wire AC is inclined at 30° to the horizontal.

Length of wire (AC):

$$\sin 30° = \frac{CE}{AC}$$

$$\frac{1}{2} = \frac{8}{AC}$$

$$AC = 16 \text{ m}$$

Distance between the poles (AE = BD):

$$\cos 30° = \frac{AE}{AC}$$

$$\frac{\sqrt{3}}{2} = \frac{AE}{16}$$

$$AE = 8\sqrt{3} \text{ m}$$

∴ The length of the wire is 16 m and the distance between the poles is $8\sqrt{3}$ m.

Source: Chapter 9 — Some Applications of Trigonometry

---

• The key step is finding the vertical difference between the two poles (28 − 20 = 8 m), which becomes the "opposite side" in the right triangle formed by the wire, the vertical height difference, and the horizontal distance.
• Use sin 30° for the wire length (hypotenuse) and cos 30° for the horizontal distance (distance between poles).
• Examiners expect a clear diagram description or labelling, correct trig ratio setup, and both answers stated explicitly.

(a) Well-labelled figure:

```
B (Top of Chhatri)
|
|
| (10–h)
|
A ————————— D
| \ 45°↑
h | \
| 60°↓ \
Water ————————————
|
C (Reflection of B)
```

• A is at height $h$ above water level.
• B is the top of Chhatri, height = 10 m above water.
• C is the reflection of B, depth = 10 m below water.
• AD is the horizontal distance. Angle of elevation ∠BAD = 45°, angle of depression ∠CAD = 60°.

---

(b) Finding h:

Let AD = horizontal distance = $d$.

From the figure:

• $BD = (10 - h)$ (B is above A by this amount)
• $CD = (10 + h)$ (C is the reflection, so 10 m below water, total vertical distance from A = $10 + h$)

From angle of elevation (∠BAD = 45°):
$$\tan 45° = \frac{10 - h}{d} \Rightarrow d = (10 - h) \quad \cdots (1)$$

From angle of depression (∠CAD = 60°):
$$\tan 60° = \frac{10 + h}{d} \Rightarrow \sqrt{3}\,d = (10 + h) \quad \cdots (2)$$

Substituting (1) in (2):
$$\sqrt{3}(10 - h) = 10 + h$$
$$10\sqrt{3} - \sqrt{3}\,h = 10 + h$$
$$10\sqrt{3} - 10 = h + \sqrt{3}\,h = h(1 + \sqrt{3})$$
$$h = \frac{10(\sqrt{3}-1)}{\sqrt{3}+1} = \frac{10(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{10(3 - 2\sqrt{3}+1)}{2} = \frac{10(4-2\times1.73)}{2}$$
$$h = \frac{10(4 - 3.46)}{2} = \frac{10 \times 0.54}{2} = \frac{5.4}{2} \approx \boxed{2.7 \text{ m}}$$

The height of point A above water level is approximately 2.7 m.

Source: Case Study — Trigonometry (Heights and Distances), Chapter 9

---

• (a) The diagram must show: point A at height h, point B (top of Chhatri) at height 10 m, point C (reflection) at depth 10 m below water, horizontal line AD, and both angles clearly marked. Examiners award marks for correct labelling and positioning.
• (b) Key insight: The reflection C is as far below water as B is above, so vertical distance from A to C = (10 + h). Set up two tan equations and solve simultaneously. Rationalise using $(\sqrt{3}-1)^2 = 4 - 2\sqrt{3}$ to get the numerical answer.

Let P be the point on the bridge, D the foot directly below P on the river bed, and A, B the two banks. Given: PD = 3 m, angle of depression to one bank = 30°, to the other = 45°.

In right △APD (∠A = 30°):
$$\tan 30° = \frac{PD}{AD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{3}{AD} \Rightarrow AD = 3\sqrt{3} \text{ m}$$

In right △BPD (∠B = 45°):
$$\tan 45° = \frac{PD}{BD} \Rightarrow 1 = \frac{3}{BD} \Rightarrow BD = 3 \text{ m}$$

Width of the river:
$$AB = AD + BD = 3\sqrt{3} + 3 = 3(\sqrt{3}+1) \text{ m}$$

Source: Chapter 9, Example 7

---

• The angle of depression from P equals the angle of elevation from the bank to P (alternate interior angles), so ∠A = 30° and ∠B = 45° inside the right triangles.
• Use tan because you know the perpendicular height (opposite side) and need the horizontal distance (adjacent side).
• The examiner expects a neat diagram description, both triangles solved separately, and the final addition clearly shown. Writing $3(\sqrt{3}+1)$ m as the answer is sufficient; no decimal approximation needed unless asked.

Let the height of the building be $h$ m and the horizontal distance between the tower and building be $d$ m.

From the foot of the building (angle of elevation of tower = 60°):
$$\tan 60° = \frac{50}{d} \implies \sqrt{3} = \frac{50}{d} \implies d = \frac{50}{\sqrt{3}}$$

From the foot of the tower (angle of elevation of building = 30°):
$$\tan 30° = \frac{h}{d} \implies \frac{1}{\sqrt{3}} = \frac{h}{d} \implies h = \frac{d}{\sqrt{3}}$$

Substituting the value of $d$:
$$h = \frac{1}{\sqrt{3}} \times \frac{50}{\sqrt{3}} = \frac{50}{3} = 16\frac{2}{3} \text{ m}$$

The height of the building is $\dfrac{50}{3}$ m $\approx 16.67$ m.

Source: Exercise 9.1, Q.9, Chapter 9

---

• Set up two right triangles sharing the same base $d$ (horizontal distance between foot of tower and foot of building).
• Use $\tan 60°$ to find $d$ from the tower's height (given = 50 m), then use $\tan 30°$ to find the building's height.
• The answer $\frac{50}{3}$ m is exact; examiners accept this or its decimal equivalent.
• Draw a neat labelled diagram — it fetches 1 mark in most CBSE marking schemes.

Setup: Let the two observers stand on the ground. Let the basket be at height $h$ m. Let the horizontal distance from the first observer (60°) to the point directly below the basket be $d$ m. The second observer is 100 m farther away, so their horizontal distance is $(d + 100)$ m.

From first observer:
$$\tan 60° = \frac{h}{d} \Rightarrow \sqrt{3} = \frac{h}{d} \Rightarrow d = \frac{h}{\sqrt{3}} \quad \cdots(1)$$

From second observer:
$$\tan 30° = \frac{h}{d+100} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{d+100} \Rightarrow d+100 = h\sqrt{3} \quad \cdots(2)$$

Substituting (1) into (2):
$$\frac{h}{\sqrt{3}} + 100 = h\sqrt{3}$$
$$100 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h\left(\frac{3-1}{\sqrt{3}}\right) = \frac{2h}{\sqrt{3}}$$
$$h = \frac{100\sqrt{3}}{2} = 50\sqrt{3} \text{ m}$$

(a) Height of the basket:
$$h = 50\sqrt{3} \approx 86.6 \text{ m}$$

(b) Distance of basket from first observer's eye (line of sight):
$$d = \frac{50\sqrt{3}}{\sqrt{3}} = 50 \text{ m}$$
$$\text{Line of sight} = \frac{h}{\sin 60°} = \frac{50\sqrt{3}}{\frac{\sqrt{3}}{2}} = 100 \text{ m}$$

(c) Horizontal distance of second observer from the basket:
$$d + 100 = 50 + 100 = 150 \text{ m}$$

Source: Chapter 9 — Heights and Distances, Section 9.1

---

• Key method: Set up two right triangles sharing the same height $h$ and solve simultaneous equations using tan of each angle.
• Examiners award marks for: correct diagram/setup (1 mark), forming both equations (1 mark), solving for $h$ (1 mark), part (b) using sin or the Pythagorean relationship for the hypotenuse (1 mark), part (c) direct substitution (1 mark).
• For (b), the "distance from the observer's eye" means the line of sight (hypotenuse), not the horizontal distance. Use $\sin 60° = h/\text{hypotenuse}$.
• Always verify: $d = 50$ m, $d+100 = 150$ m, and check $\tan 30° = 50\sqrt{3}/150 = 1/\sqrt{3}$ ✓

Let AB = 7 m (building), CD = height of cable tower, BD = horizontal distance between them.

Draw AE ∥ BD, so AE = BD and ED = AB = 7 m.

Step 1: Find BD using angle of depression (30°)

From A, angle of depression of foot D = 30°.
In △ABD, ∠ADB = 30° (alternate angles):

$$\tan 30° = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{7}{BD} \implies BD = 7\sqrt{3} \text{ m}$$

So AE = $7\sqrt{3}$ m.

Step 2: Find CE using angle of elevation (60°)

From A, angle of elevation of top C = 60°.
In △AEC:

$$\tan 60° = \frac{CE}{AE} \implies \sqrt{3} = \frac{CE}{7\sqrt{3}} \implies CE = 7\sqrt{3} \times \sqrt{3} = 21 \text{ m}$$

Step 3: Total height of tower

$$CD = CE + ED = 21 + 7 = \boxed{28 \text{ m}}$$

The height of the cable tower is 28 m.

Source: Chapter 9, Section 9.1 (Heights and Distances)

---

• The key construction is drawing AE horizontal (parallel to BD), splitting the tower height into two parts: ED = 7 m (equal to building height) and CE (found using tan 60°).
• Angle of depression = 30° gives the horizontal distance BD via tan 30°; this same distance is AE used in the second triangle.
• Examiners award marks for: correct diagram (1), correct use of tan 30° to find BD (1–2), correct use of tan 60° to find CE (1–2), and final addition (1). Always show both triangles separately and write the final answer clearly.
• Note: The textbook (Q.12 in Exercise 9.1) has 45° for angle of depression, giving a different answer (7+7√3 m). This question uses 30°, so the answer is 28 m.

Let AB = 75 m be the tower. Let C and D be the two cars on the highway, with D farther from the tower.

Diagram: Tower AB vertical; C and D on the ground with D beyond C from the tower. Angles of depression to C and D are 60° and 30° respectively.

In right △ABC (angle of depression = 60°):

$$\tan 60° = \frac{AB}{BC}$$

$$\sqrt{3} = \frac{75}{BC}$$

$$BC = \frac{75}{\sqrt{3}} = \frac{75\sqrt{3}}{3} = 25\sqrt{3} \text{ m}$$

In right △ABD (angle of depression = 30°):

$$\tan 30° = \frac{AB}{BD}$$

$$\frac{1}{\sqrt{3}} = \frac{75}{BD}$$

$$BD = 75\sqrt{3} \text{ m}$$

Distance between the two cars:

$$CD = BD - BC = 75\sqrt{3} - 25\sqrt{3} = 50\sqrt{3} \text{ m}$$

$$CD = 50 \times 1.73 = \textbf{86.5 m}$$

Source: Chapter 9 (Some Applications of Trigonometry), Exercise 9.1

---

• Key idea: Angle of depression from the top = angle of elevation from the car to the top (alternate angles with the horizontal). So use tan directly in the right triangle formed by tower height and ground distance.
• The car at 60° is closer (smaller distance BC); the car at 30° is farther (larger distance BD). Distance between cars = BD − BC.
• Always rationalize $\frac{75}{\sqrt{3}}$ to get $25\sqrt{3}$.
• Substitute $\sqrt{3} = 1.73$ only at the final step.
• Draw and label the diagram — it typically carries 1 mark in CBSE.

(a) 60°

Here, pole height = 6 m, shadow length = $2\sqrt{3}$ m. $\tan\theta = \dfrac{6}{2\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$, so $\theta = 60°$.

Set up a right triangle where the pole is the opposite side and the shadow is the adjacent side. Using $\tan\theta = \frac{\text{height}}{\text{shadow}}$ gives $\sqrt{3}$, which corresponds to 60°. Remember: $\tan 60° = \sqrt{3}$, $\tan 45° = 1$, $\tan 30° = \frac{1}{\sqrt{3}}$.

Diagram: Let P be the point on the ground, AB = 30 m be the building, and BC = h be the transmission tower. Let PA = d.

Step 1: In △PAB (angle of elevation to bottom of tower = 30°):

$$\tan 30° = \frac{AB}{PA} \Rightarrow \frac{1}{\sqrt{3}} = \frac{30}{d}$$

$$d = 30\sqrt{3} \text{ m}$$

Step 2: In △PAC (angle of elevation to top of tower = 60°, AC = AB + BC = 30 + h):

$$\tan 60° = \frac{AC}{PA} \Rightarrow \sqrt{3} = \frac{30 + h}{30\sqrt{3}}$$

$$30 + h = \sqrt{3} \times 30\sqrt{3} = 90$$

$$h = 90 - 30 = 60 \text{ m}$$

Therefore, the height of the transmission tower = 60 m.

(Using $\sqrt{3} = 1.73$, the distances are consistent; the tower height is exactly 60 m.)

Source: Chapter 9, Exercise 9.1 (Q.7 variant), Heights and Distances

---

• The key is setting up two separate right triangles from the same point P: one to the base of the tower (bottom of tower = top of building) and one to the top of the tower.
• Use tan for both since you have opposite (height) and adjacent (horizontal distance) sides.
• Find the horizontal distance from the first triangle, then substitute into the second.
• Examiners award marks for: correct diagram (1 mark), correct equation for tan 30° (1 mark), correct equation for tan 60° (1 mark), solving for d (1 mark), and final answer h = 60 m (1 mark).

Diagram: Let AB = 75 m be the lighthouse. Let C and D be the two ships, with D farther from the lighthouse. Angles of depression to D and C are 30° and 60° respectively.

In right △ABC (nearer ship, angle of depression = 60°):

$$\tan 60° = \frac{AB}{BC}$$

$$\sqrt{3} = \frac{75}{BC} \implies BC = \frac{75}{\sqrt{3}} = 25\sqrt{3} \text{ m}$$

In right △ABD (farther ship, angle of depression = 30°):

$$\tan 30° = \frac{AB}{BD}$$

$$\frac{1}{\sqrt{3}} = \frac{75}{BD} \implies BD = 75\sqrt{3} \text{ m}$$

Distance between the two ships:

$$CD = BD - BC = 75\sqrt{3} - 25\sqrt{3} = 50\sqrt{3} \text{ m}$$

$$= 50 \times 1.73 = \boxed{86.5 \text{ m}}$$

Source: Chapter 9, Heights and Distances (Exercise 9.1, Q.13)

---

• The angle of depression from the top of the lighthouse equals the angle of elevation from the ship to the top (alternate angles), so tan is applied directly with height = 75 m.
• The ship closer to the lighthouse makes a larger angle of depression (60°), and the farther ship makes a smaller angle (30°) — a common point students get wrong.
• CD = BD − BC is the key final step.
• Always substitute √3 = 1.73 at the end and state the unit (metres).

Let AB be the height of the tower and BC = 30 m be the distance from the foot.

In right △ABC, ∠ACB = 30°

$$\tan 30° = \frac{AB}{BC}$$

$$\frac{1}{\sqrt{3}} = \frac{AB}{30}$$

$$AB = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ m}$$

∴ The height of the tower is $10\sqrt{3}$ m.

Source: Exercise 9.1, Q.4; Chapter 9 — Some Applications of Trigonometry

---

• Draw a right triangle with the tower as the vertical side (AB), the ground distance as the base (BC = 30 m), and angle of elevation at C = 30°.
• Use tan because you have the opposite side (height) and adjacent side (distance).
• Rationalise $\frac{30}{\sqrt{3}}$ by multiplying numerator and denominator by $\sqrt{3}$ to get $10\sqrt{3}$.
• Examiner expects: diagram description (optional but good), correct trig ratio, clear working, and the final answer in simplified surd form.

Let the height of the tower be $h$ and the length of its shadow be $\sqrt{3}\,h$.

Let $\theta$ be the angle of elevation of the sun.

In the right triangle formed:

$$\tan\theta = \frac{\text{height of tower}}{\text{length of shadow}} = \frac{h}{\sqrt{3}\,h} = \frac{1}{\sqrt{3}}$$

Since $\tan 30° = \dfrac{1}{\sqrt{3}}$, we get $\theta = 30°$.

The angle of elevation of the sun is 30°.

Source: Chapter 9, Heights and Distances (Exercise 9.1)

---

• Draw the right triangle with the tower as the perpendicular side and the shadow as the base.
• Use $\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{\sqrt{3}h}$; the $h$ cancels, leaving $\frac{1}{\sqrt{3}}$.
• Recall the standard value: $\tan 30° = \frac{1}{\sqrt{3}}$, so $\theta = 30°$.
• Always state the final answer clearly; examiners award the last mark for the correct conclusion.

Diagram: Let P be the point on the bridge, D the foot of the vertical from P to the river level. Let A and B be the two banks, with angles of depression 30° and 60° respectively. PD = 4 m.

AB = AD + DB is the width of the river.

In right △APD (angle of depression = 30°, so ∠PAD = 30°):

$$\tan 30° = \frac{PD}{AD}$$

$$\frac{1}{\sqrt{3}} = \frac{4}{AD}$$

$$AD = 4\sqrt{3} \text{ m}$$

In right △PBD (angle of depression = 60°, so ∠PBD = 60°):

$$\tan 60° = \frac{PD}{BD}$$

$$\sqrt{3} = \frac{4}{BD}$$

$$BD = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \text{ m}$$

Width of the river:

$$AB = AD + BD = 4\sqrt{3} + \frac{4\sqrt{3}}{3} = \frac{12\sqrt{3} + 4\sqrt{3}}{3} = \frac{16\sqrt{3}}{3} \text{ m}$$

$$\boxed{AB = \dfrac{16\sqrt{3}}{3} \text{ m}}$$

Source: Chapter 9, Section 9.1 Heights and Distances (cf. Example 7)

---

• This is a direct application of Example 7 in the textbook, but with height 4 m and angles 30° and 60° (instead of 3 m, 30° and 45°).
• Key step: the angle of depression from the bridge equals the angle of elevation from the bank (alternate interior angles), so use tan directly in each right triangle.
• Don't forget to rationalize $\frac{4}{\sqrt{3}}$ to get $\frac{4\sqrt{3}}{3}$, then add the two distances carefully.
• Examiners award marks for the diagram, correct trig equations, correct individual distances, and the final addition — show each step.

Diagram: Let AB and CD be two pillars of equal height h m. Let P be the point on the road between them, with BP = x m, so PD = (100 − x) m.

In △ABP (angle of elevation = 60°):
$$\tan 60° = \frac{AB}{BP} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \cdots (1)$$

In △CDP (angle of elevation = 30°):
$$\tan 30° = \frac{CD}{PD} \implies \frac{1}{\sqrt{3}} = \frac{h}{100 - x} \implies h = \frac{100 - x}{\sqrt{3}} \quad \cdots (2)$$

From (1) and (2):
$$x\sqrt{3} = \frac{100 - x}{\sqrt{3}}$$
$$3x = 100 - x \implies 4x = 100 \implies x = 25 \text{ m}$$

Height of each pillar:
$$h = 25\sqrt{3} = 25 \times 1.732 = 43.3 \text{ m}$$

Distances:

• Distance from pillar AB = BP = 25 m
• Distance from pillar CD = PD = 100 − 25 = 75 m

∴ The length of each pillar is 43.3 m, and the point P is 25 m from one pillar and 75 m from the other.

Source: Chapter 9, Exercise 9.1 (Q.10 variant), Heights and Distances

---

• Set up two right triangles sharing the point P; label distances as x and (100 − x).
• Use tan for both elevation angles since you have opposite (height) and adjacent (horizontal distance).
• Equate the two expressions for h to find x, then substitute back.
• Examiners award marks for: diagram/setup (1), correct equations (1+1), solving for x (1), final height with √3 value (1).
• Note: the pillar with the larger elevation angle (60°) is closer to P — a quick sanity check.

Option B: 45°

When shadow length = height of tower, tan θ = height/shadow = 1, so θ = 45°.

Let tower height = h; shadow length = h. Sun's altitude = angle of elevation θ, so tan θ = h/h = 1 ⟹ θ = 45°. Remember: tan 45° = 1 is a standard trigonometric value.

Diagram: Let AB = 20 m (building), BC = tower, P = point on ground, PB = horizontal distance.

Step 1: In △PAB, angle of elevation of bottom of tower (top of building) = 45°.

$$\tan 45° = \frac{AB}{PA} \implies 1 = \frac{20}{PA} \implies PA = 20 \text{ m}$$

Step 2: Let height of tower BC = h m. Then total height PC = (20 + h) m.

In △PAC, angle of elevation of top of tower = 60°.

$$\tan 60° = \frac{AC}{PA} \implies \sqrt{3} = \frac{20 + h}{20}$$

Step 3: Solving:

$$20\sqrt{3} = 20 + h$$

$$h = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1) \text{ m}$$

∴ Height of the tower = $20(\sqrt{3} - 1)$ m ≈ 14.64 m

Source: Exercise 9.1, Q.7; Chapter 9 — Some Applications of Trigonometry

---

• The key is setting up two separate right triangles from the same point P: one to the base of the tower (top of building) and one to the top of the tower.
• Since 45° elevation gives tan 45° = 1, the horizontal distance equals the building height (20 m) — this is the crucial first step.
• Examiners expect a labelled diagram, both tan equations shown clearly, and the final answer simplified to $20(\sqrt{3}-1)$ m. Writing the decimal approximation is optional but appreciated.

Option B: $30\sqrt{3}$ m

Using $\tan 60° = \dfrac{h}{30}$, we get $\sqrt{3} = \dfrac{h}{30}$, so $h = 30\sqrt{3}$ m.

This directly applies $\tan(\text{angle of elevation}) = \dfrac{\text{height}}{\text{base distance}}$. The base is 30 m and angle is 60°, so multiply 30 by $\tan 60° = \sqrt{3}$. This mirrors Example 1 of Chapter 9 (where base = 15 m gave $15\sqrt{3}$). Don't confuse with Exercise Q4 (which has the same base but angle 30°, giving a different answer).

Diagram: Let AB = 45 m be the lighthouse. Ships are at C and D on opposite sides, with foot of lighthouse at B.
Angle of depression to C = 60°, angle of depression to D = 30°.

For Ship C (angle of depression = 60°):

In △ABC,
$$\tan 60° = \frac{AB}{BC}$$
$$\sqrt{3} = \frac{45}{BC}$$
$$BC = \frac{45}{\sqrt{3}} = \frac{45\sqrt{3}}{3} = 15\sqrt{3} \text{ m}$$

For Ship D (angle of depression = 30°):

In △ABD,
$$\tan 30° = \frac{AB}{BD}$$
$$\frac{1}{\sqrt{3}} = \frac{45}{BD}$$
$$BD = 45\sqrt{3} \text{ m}$$

Distance between the two ships:
$$CD = BC + BD = 15\sqrt{3} + 45\sqrt{3} = 60\sqrt{3} \text{ m}$$
$$= 60 \times 1.73 = \boxed{103.8 \text{ m}}$$

Source: Some Applications of Trigonometry, Chapter 9

---

• Since the ships are on opposite sides, distances BC and BD are added (not subtracted).
• The angle of depression from the top equals the angle of elevation from the ship's position (alternate angles with horizontal), so directly use tan in the right triangle formed.
• Always rationalize $\frac{45}{\sqrt{3}}$ to get $15\sqrt{3}$ — examiners check this step.
• Substitute $\sqrt{3} = 1.73$ only in the final step to avoid rounding errors midway.
• Drawing a neat labelled diagram typically earns 1 mark on its own.

Let A be the top of the lighthouse, B be the base, height AB = 50 m.

(i) Distance of ship at Q (angle of depression = 45°):

In △ABQ: tan 45° = AB/BQ
$$1 = \frac{50}{BQ} \Rightarrow BQ = 50 \text{ m}$$

(ii) Measures of ∠PBA and ∠QBA:

Since AB ⊥ BP (base is horizontal), both ∠PBA and ∠QBA = 90°.

(iii) Distance travelled by the ship (PQ):

In △ABP: tan 30° = AB/BP
$$\frac{1}{\sqrt{3}} = \frac{50}{BP} \Rightarrow BP = 50\sqrt{3} \text{ m}$$

Distance travelled = PQ = BP − BQ
$$= 50\sqrt{3} - 50 = 50(\sqrt{3}-1) \text{ m} \approx 50(1.732-1) = 36.6 \text{ m}$$

Source: Chapter 9 – Some Applications of Trigonometry

---

• Part (i): Straightforward tan ratio with 45°; tan 45° = 1 so BQ = height = 50 m.
• Part (ii): ∠PBA and ∠QBA are the angles at base B in the respective right triangles, both equal 90° since the lighthouse is vertical and the ground is horizontal.
• Part (iii): Find BP using tan 30°, then subtract BQ from BP. Remember to simplify and rationalize. Examiners expect the expression $50(\sqrt{3}-1)$ m clearly shown.

(C) Assertion (A) is true, but Reason (R) is false.

Assertion is true: by Pythagoras theorem, ladder = $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ m. Reason is false: the angle the ladder makes with the ground is $\tan^{-1}(8/6) \approx 53°$, not $60°$.

• Verify the Assertion using the Pythagoras theorem (no trigonometry needed).
• To check Reason, use $\tan\theta = \text{opposite}/\text{adjacent} = 8/6 = 4/3$; this gives $\approx 53°$, not $60°$. So Reason is false, making option (C) correct.
• In Assertion-Reason questions, always verify both statements independently before choosing the option.

Diagram: Let PC = multi-storeyed building, AB = 8 m tall building. Draw PQ horizontal from P. Angles of depression to top (B) and bottom (A) of AB are 30° and 45° respectively.

By alternate angles: ∠PBD = 30° and ∠PAC = 45°, where BD ∥ AC (horizontal distances).

Let PD = h m (height above B), so PC = PD + DC = h + 8.
Also, AC = BD (distance between buildings).

In right △PAC:
$$\tan 45° = \frac{PC}{AC} \Rightarrow 1 = \frac{PC}{AC} \Rightarrow AC = PC$$

In right △PBD:
$$\tan 30° = \frac{PD}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{BD} \Rightarrow BD = h\sqrt{3}$$

Since AC = BD and PC = h + 8:
$$h + 8 = h\sqrt{3}$$
$$h(\sqrt{3}-1) = 8 \Rightarrow h = \frac{8}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = 4(\sqrt{3}+1) \text{ m}$$

Height of multi-storeyed building:
$$PC = 4(\sqrt{3}+1)+8 = 4\sqrt{3}+4+8 = 4(\sqrt{3}+3) \text{ m}$$

Distance between buildings:
$$AC = PC = 4(\sqrt{3}+3) \text{ m}$$

Source: Chapter 9 – Heights and Distances, Example 6

---

• Draw a clear labelled diagram — examiners award 1 mark for it.
• The key step is using alternate interior angles to convert angles of depression into angles of elevation inside the right triangles (∠QPB = ∠PBD = 30°, etc.).
• Set up two tan equations, then solve simultaneously using AC = BD and PC = PD + 8.
• Rationalise the denominator when computing PD — leave answer in surd form as $4(\sqrt{3}+3)$ m.
• Both the height and the distance come out equal here, which is a good self-check.

Let the height of the lighthouse = h m.

Let the lighthouse be AB, with ships at C and D on either side.

• Angle of depression to ship C = 60° → ∠ACB = 60° (alternate angles)
• Angle of depression to ship D = 45° → ∠ADB = 45°

In right △ABC (angle 60°):
$$\tan 60° = \frac{h}{BC} \Rightarrow \sqrt{3} = \frac{h}{BC} \Rightarrow BC = \frac{h}{\sqrt{3}}$$

In right △ABD (angle 45°):
$$\tan 45° = \frac{h}{BD} \Rightarrow 1 = \frac{h}{BD} \Rightarrow BD = h$$

Total distance between ships:
$$BC + BD = \frac{h}{\sqrt{3}} + h = h\left(\frac{1}{\sqrt{3}} + 1\right) = h \cdot \frac{\sqrt{3}+1}{\sqrt{3}}$$

Given: $BC + BD = 100\left(\dfrac{\sqrt{3}+1}{\sqrt{3}}\right)$

$$h \cdot \frac{\sqrt{3}+1}{\sqrt{3}} = 100 \cdot \frac{\sqrt{3}+1}{\sqrt{3}}$$

$$\boxed{h = 100 \text{ m}}$$

The height of the lighthouse is 100 m.

Source: Some Applications of Trigonometry, Chapter 9

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• Key concept: Angle of depression from the top equals the angle of elevation from the ship's position (alternate interior angles with the horizontal). Use tan ratio in each right triangle.
• Common mistake: Forgetting that the ships are on opposite sides, so their distances add up (not subtract).
• Examiner expects: Diagram/setup (1 mark), two tan equations (2 marks), sum = given distance (1 mark), final answer (1 mark). Always state the conclusion clearly.

Let CE = height of lighthouse, CD = horizontal distance, DE = h (height above D), so CE = CD + DE...

Setup: Let BC = h (height of lighthouse above observation deck level), so BE = h. Let CD = horizontal distance from Amrita to lighthouse base.

(i) From point B (observation deck), angle of elevation = 45°:
$$\tan 45° = \frac{BE}{BD} \Rightarrow 1 = \frac{h}{BD} \Rightarrow BD = h$$
$$\therefore CD = BD = h$$

(ii) From point B, angle of elevation = 45°:
$$\tan 45° = \frac{BE}{BD} = 1 \Rightarrow BD = BE = BC$$
$$\therefore BD = BC$$

(iii) From point A (ground level), angle of elevation = 60°. AB = 40 m, so AE = h + 40:
$$\tan 60° = \frac{h + 40}{CD} \Rightarrow \sqrt{3} = \frac{h + 40}{h}$$
$$\sqrt{3}\,h = h + 40 \Rightarrow h(\sqrt{3} - 1) = 40$$
$$h = \frac{40}{\sqrt{3}-1} = \frac{40(\sqrt{3}+1)}{2} = 20(\sqrt{3}+1)$$
$$h = 20(1.73 + 1) = 20 \times 2.73 = 54.6 \text{ m}$$
$$\therefore CE = h + 40 = 54.6 + 40 = \textbf{94.6 m}$$

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• The key is setting BC = h (lighthouse height above deck level B) and noting CD = BD since tan 45° = 1.
• For part (iii), from A the full height seen is (h + 40) over horizontal distance h, giving the equation with √3.
• Rationalize the denominator properly. Examiners check the rationalization step and correct substitution of √3 = 1.73.
• Label all parts clearly — examiners award step marks for correct tan ratio setup, equation, and final answer.

Option B: 300 m

Using $\sin 30° = \dfrac{\text{height}}{\text{string length}}$, we get $\dfrac{1}{2} = \dfrac{150}{L}$, so $L = 300$ m.

The height (150 m) is the side opposite the 30° angle, and the string is the hypotenuse. Use $\sin\theta = \text{opposite}/\text{hypotenuse}$. $\sin 30° = \frac{1}{2}$, giving $L = 150 \times 2 = 300$ m. Don't confuse with $\tan$ or $\cos$ here.

Diagram: Let the ground be at point A. The kite is at point K, 60 m above the ground. The building stands vertically; roof is at point R, ground level at point B (foot of building).

Let height of roof = h m. Height of kite above roof = (60 − h) m.
Horizontal distance between kite and building = d m.

From the bottom of the building (point B), angle of elevation = 45°:

$$\tan 45° = \frac{60}{d}$$
$$1 = \frac{60}{d} \implies d = 60 \text{ m}$$

From the roof (point R), angle of elevation = 30°:

$$\tan 30° = \frac{60 - h}{d}$$
$$\frac{1}{\sqrt{3}} = \frac{60 - h}{60}$$
$$60 - h = \frac{60}{\sqrt{3}} = 20\sqrt{3}$$
$$h = 60 - 20\sqrt{3} = 60 - 20 \times 1.73 = 60 - 34.6 = 25.4 \text{ m}$$

Length of string (from roof R to kite K):

$$\sin 30° = \frac{60 - h}{\text{string}}$$
$$\frac{1}{2} = \frac{20\sqrt{3}}{\text{string}}$$
$$\text{String} = 40\sqrt{3} = 40 \times 1.73 = 69.2 \text{ m}$$

∴ Height of roof = 25.4 m; Length of string = 69.2 m.

Source: Chapter 9 — Heights and Distances, Section 9.1

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• Two right triangles are formed: one from the ground (angle 45°) and one from the roof (angle 30°), both looking at the same kite at 60 m height.
• From the ground triangle, find horizontal distance first (tan 45° = 1 makes it straightforward).
• Use that distance in the roof triangle to find the roof height.
• For the string length, use sin 30° with the vertical height difference (60 − h) as the opposite side — the string is the hypotenuse from Ravi's position to the kite.
• Always substitute √3 = 1.73 numerically at the end as the question requires.