# Applications of Trigonometry Heights and Distances Chapter 9 Exercise 9.1 Question 6

Applications of Trigonometry Heights and Distances Chapter 9 Exercise 9.1 Question 6

6.    A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30$^\circ$ to 60$^\circ$ as he walks towards the building. Find the distance he walked towards the building.

Solution:

We are given the height of tower (AF) = 30 m

The height of boy (BD) = 1.5 m

Therefore, AC = AF - BD = 30 -1.5 = 28.5 m

We want to find distance he walked towards the tower which is BG.

In $\triangle AGC$, we have $\frac{GC}{AC}= cot 60^\circ$

$\Rightarrow GC = 28.5 \times \frac{1}{\sqrt{3}} =\frac{28.5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow GC = 9.5\sqrt{3}$ m

In $\triangle ABC$, we have $\frac{BC}{AC}= cot 30^\circ$

$\Rightarrow BC = 28.5 \times \sqrt{3} = 28.5\sqrt{3}$ m

BG=BC-GC $=28.5\sqrt{3} -9.5\sqrt{3}=19\sqrt{3}$ m

Therefore, he walks $19\sqrt{3}$ m towards the tower.

Posted in : NCERT Solutions, Trigonometry

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## 2 thoughts on “Applications of Trigonometry Heights and Distances Chapter 9 Exercise 9.1 Question 6”

1. Jashan Post author

As far as I understand your question, are you trying to ask how do we get to know that we are supposed to use sin, cos or tan. It all depends on which sides are given to us.

Thanks