Question

A boy $1.6\,{\text{m}}$ tall, is $20\,{\text{m}}$ away from a tower and observes the angle of elevation of the top of the tower to be (i) ${45^ \circ }$ (ii) ${60^ \circ }$ . Find the height of the tower in each case.

Answer 1

Hint : Draw a diagram using the given parameters. Observe the diagram carefully and using trigonometric identities try to find the height of the tower. There are two angles of elevation given so find the height of the tower separately for both angles.

Complete step-by-step answer :
Given, height of the boy ,
Distance between the boy and the tower, $d = 20\,{\text{m}}$
We are asked to find the height of the tower for two angles of elevation.

(i) ${45^ \circ }$
Let us first draw a diagram

Here, DC is the height of the tower,
$AB = 20\,{\text{m}}$ , $BD = 1.6\;{\text{m}}$
$\angle CAB = {45^ \circ }$
Triangle ABC forms a right angled triangle, in the right angle triangle we have, $\tan \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$
And here for angle $\angle CAB$ the perpendicular is BC and base is AB, so we can write,
$\tan {45^ \circ } = \dfrac{{BC}}{{AB}}$

\Rightarrow BC = AB $$ Putting the value of AB we get, $$BC = 20\;{\text{m}}$$ Height of the tower is, $$DC = BD + BC$$ Putting the values of BD and BC we get, $$DC = 1.6 + 20 = 21.6\;{\text{m}}$$ Therefore, the height of the tower is $$21.6\;{\text{m}}$$ . **So, the correct answer is “ $$21.6\;{\text{m}}$$ ”.** (ii) $${60^ \circ }$$  Here, $$AB = 20\,{\text{m}}$$ , $$BD = 1.6\;{\text{m}}$$ $$\angle CAB = {60^ \circ }$$ We will use the formula for $$\tan \theta $$ as we had used for part (i). Here, $$\tan {60^ \circ } = \dfrac{{BC}}{{AB}}$$ $$ \Rightarrow BC = AB\tan {60^ \circ } = AB\sqrt 3 $$ Putting the value of AB we get, $$BC = 20\sqrt 3 {\text{m}}$$ Height of the tower is, $$DC = BD + BC$$ Putting the values of BD and BC we get, $$DC = \left( {1.6 + 20\sqrt 3 } \right){\text{m}}$$ Therefore, the height of the tower is $$1.6 + 20\sqrt 3 \;{\text{m}}$$ . **So, the correct answer is “ $$1.6 + 20\sqrt 3 \;{\text{m}}$$ ”.** **Note** : Angle of elevation is the angle between the horizontal line and line of sight. Here, line of sight can be said as an imaginary line drawn from the boy’s eyes to the top of the tower. Whenever such questions are given, use the trigonometric identities for sine, cosine and tangent to find the height of an object or distance between the object and the body.