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

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

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• 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.