Mathematics — CBSE Class 10 board question

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

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• 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; $\sqrt{448} = \sqrt{64 \times 7} = 8\sqrt{7}$.