Question

The angles of depression of the top and the bottom of a 10 m tall building from the top of a multi-storeyed building are ${30}^{\circ }$ and ${45}^{\circ }$, respectively. Find the height of the multi-storeyed building.

• A. 10 m
• B. 15 m
• C. 5($\sqrt{3}$ + 3) m
• D. 5 m

Answer 1

The correct option is C

5($\sqrt{3}$ + 3) m

The above figure represents the situation aptly
$\angle CBE=\angle BEFand\angle DAE=\angle AEF\left(\text{alternate angles}\right)tan\left(\angle EAD\right)=tan{45}^{\circ }=\frac{ED}{AD}=\frac{EC+CD}{AD}\Rightarrow AD\times tan{45}^{\circ }=EC+CD\dots \dots \left(1\right)andtan\left(\angle EBC\right)=tan{30}^{\circ }=\frac{EC}{CB}\Rightarrow CB\times tan{30}^{\circ }=EC\dots \dots \left(2\right)$
Subtracting eq(2) from (1)
$\Rightarrow AD\times tan{45}^{\circ }-CB\times tan{30}^{\circ }=CD\Rightarrow AD(tan{45}^{\circ }-tan{30}^{\circ })=CD(\because AD=CB)\Rightarrow AD(1-\frac{1}{\sqrt{3}})=10\Rightarrow AD=5(3+\sqrt{3})m\Rightarrow ED=AD(tan{45}^{\circ }=1)$

ED is the height of the building.