**Question:**

From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find height of the tower.

**Solution:**

It is given that tower is placed at a 20 m high building. The top and bottom of the tower makes an angle of respectively with the ground.

We have to find the height of the tower.

Let *DB* is the tower

Let *AD* is the building

Height of the building = 20 m

Height of the tower = *x *m

According to the figure,

$\frac{A B}{A C}=\tan 60^{\circ}$

$A C=\frac{A B}{\tan 60^{\circ}}$.......(1)

$\frac{A D}{A C}=\tan 45^{\circ}$

$A C=\frac{A D}{\tan 45^{\circ}} \ldots \ldots(2)$

Since (1) = (2)

$\frac{A B}{\tan 60^{\circ}}=\frac{A D}{\tan 45^{\circ}}$

$\frac{20+x}{\sqrt{3}}=\frac{20}{1}$

$20+x=20 \sqrt{3}$

$x=20 \sqrt{3}-20$

$x=20(\sqrt{3}-1)$

$x=20 \times 0.732$

$x=14.64 \mathrm{~m}$

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