Mathematics — Standard (041) — CBSE Class 10 board question

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

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• 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.