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 \(\begin{align}C D=16 \frac{2}{3} \mathrm{m}\end{align}\)