The angles of depression of the top and bottom of a 20 m high building from the top of a - Maths -

Question

The angles of depression of the top and bottom of a 20 m high
building from the top of a multistoryed building are 300
and 450 respectively Find the height of the multistoryed
building (Take root 3 = 1 732)

Shridhar Kaushik · Accepted Answer

Let CD be the building with CD = 20m and AB = h m be the height of multistoryed building.

∠CAE = 30° and ∠CBD = 45°

As AB = h m ⇒ CE = (20 – h) m

So in right angled ∆ACE,

$\tan 30 ° = \frac{CE}{AE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{20 - h}{AE} \Rightarrow AE = \sqrt{3} (20 - h) ............ (1)$

In ∆BCD,

$\tan 45 ° = \frac{CD}{BD} \Rightarrow CD = BD \cdot \tan 45 °$

⇒CD = BD·1

⇒BD = CD ............(2)

as BD = AE, so from (1) and (2)

$CD = \sqrt{3} (20 - h)$

Given CD = 20m

$∴ 20 = \sqrt{3} \times 20 - \sqrt{3} h \Rightarrow \sqrt{3} h = \sqrt{3} \times (20) - 20 \Rightarrow \sqrt{3} h = 20 (\sqrt{3} - 1) \Rightarrow \sqrt{3} h = 20 (1.732 - 1) \Rightarrow h = \frac{20 \times 0.732}{1.732} m$

⇒ h = 8.45 m

Hence, the height of multistoryed building is 8.45 m.

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The angles of depression of the top and bottom of a 20 m high building from the top of a multistoryed building are 300 and 450 respectively. Find the height of the multistoryed building. (Take root 3 = 1.732)

The angles of depression of the top and bottom of a 20 m high building from the top of a multistoryed building are 30⁰ and 45⁰ respectively. Find the height of the multistoryed building. (Take root 3 = 1.732)