# Ex.9.1 Q9 Some Applications of Trigonometry Solution - NCERT Maths Class 10

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## Question

The angle of elevation of the top of a building from the foot of the tower is $$30^\circ$$ and the angle of elevation of the top of the tower from the foot of the building is $$60^\circ.$$ If the tower is $$50\,\rm{m}$$ high, find the height of the building.

## Text Solution

What is Known?

(i) Angle of elevation of the top of a building from the foot of the tower  $$=30^\circ$$

(ii) Angle of elevation of the top of the tower from the foot of the building $$=60^\circ$$

(iii) Height of tower $$= 50\,{\rm{m}}$$

What is  Unknown?

Height of the building Reasoning:

Let the height of the tower is $$AB$$ and the height of the building is $$CD$$. The angle of elevation of the top of the building $$D$$ from the foot of the tower $$B$$ is $$30^\circ$$ and the angle of elevation of the top of the tower $$A$$ from the foot of the building $$C$$ is $$60^\circ$$.

Distance between the foot of the tower and the building is $$BC$$.

Trigonometric ratio involving sides $$AB, \,CD, \,BC$$ and angles $$∠B$$ and $$∠C$$ is $$\tan \theta$$

Steps:

In  $$\Delta ABC$$

\begin{align}{\text{tan 6}}{{\rm{0}}^0} &= \frac{{AB}}{{BC}}\\\sqrt 3 &= \frac{{50}}{{BC}}\\BC &= \frac{{50}}{{\sqrt 3 }}\qquad\left( {\rm{i}} \right)\end{align}

In $$\Delta BCD$$,

\begin{align} \tan {{30}^{0}}&=\frac{CD}{BC} \\ \frac{1}{\sqrt{3}}&=\frac{CD}{BC} \\ \frac{1}{\sqrt{3}}&=\frac{CD}{\frac{50}{\sqrt{3}}}\qquad\left[ \text{from}\left( \text{i} \right) \right] \\ CD&=\frac{1}{\sqrt{3}}\times \frac{50}{\sqrt{3}} \\ CD&=\frac{50}{3} \\ CD&=16\frac{2}{3} \end{align}

Height of the building \begin{align}C D=16 \frac{2}{3} \mathrm{m}\end{align}

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