Question

15. The angle of elevation of the top of a pillar from
a point on the ground is 15° on walking 100 m
towards the tower, the angle of elevation is
found to be 30°. Calculate the height of the tower
(where tan 15° = 2 -V3).​

Answer 1

Solution :

$\underline{\bf{Given\::}}}$

The angle of elevation of the top of a pillar from a point on the ground is 15° on walking 100 m towards the tower, the angle of elevation if found to be 30°.

$\underline{\bf{Explanation\::}}}$

Attachment a figure according to the question;

• Walking on the ground, (AB) = 100 m
• BC = r m

Now;

$\underline{\sf{In\: \triangle BCD \:where \:\angle \:C = 90\degree\::}}$

$\longrightarrow\sf{tan\theta=\dfrac{Perpendicular}{Base} }$

$\longrightarrow\sf{tan30\degree=\dfrac{CD}{BC} }\\\\\\\longrightarrow\sf{\dfrac{1}{\sqrt{3} } =\dfrac{h}{r} \:\:[\therefore tan30\degree = 1/\sqrt{3} ]}\\\\\\\longrightarrow\sf{r=\sqrt{3} h\:m..............(1)}$

$\underline{\sf{In\: \triangle ACD \:where \:\angle \:C = 90\degree\::}}$

$\longrightarrow\sf{tan15\degree = \dfrac{DC}{AC} }\\\\\\\longrightarrow\sf{2-\sqrt{3} = \dfrac{h}{AB+BC} }\\\\\\\longrightarrow\sf{2-\sqrt{3} =\dfrac{h}{100+\sqrt{3}h } }\\\\\\\longrightarrow\sf{h=2-\sqrt{3} (100+\sqrt{3} h)}\\\\\\\longrightarrow\sf{100+\sqrt{3} h=\dfrac{h}{2-\sqrt{3} } }\\\\\\\longrightarrow\sf{100+\sqrt{3} h=\dfrac{h\times 2+\sqrt{3}}{2-\sqrt{3} \times 2+\sqrt{3}} \:\:[Rationalization]}\\\\\\\longrightarrow\sf{100+\sqrt{3} h=\dfrac{h(2+\sqrt{3} )}{4 -3} }$

$\longrightarrow\sf{100+\sqrt{3} h=\dfrac{h(2+\sqrt{3} )}{1} }\\\\\longrightarrow\sf{100+\sqrt{3} h=h(2+\sqrt{3} )}\\\\\longrightarrow\sf{100\cancel{+\sqrt{3} h} = 2h \cancel{+\sqrt{3} h}}\\\\\longrightarrow\sf{100=2h}\\\\\longrightarrow\sf{h=\cancel{100/2}}\\\\\longrightarrow\bf{h=50\:m}$

Thus;

The height of the tower will be 50 m .