JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Heights and Distances

From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is

6 0^{\circ}

. The pole subtends an angle

3 0^{\circ}

at the top of the tower. Then the height of the tower is:

Answer: (D)

Solution

Let the tower have height

h

and let the horizontal distance from the pole to the tower be

d

.

From the base of the pole, the angle of elevation to the top of the tower is

6 0^{\circ}

tan 6 0^{\circ} = \frac{h}{d} \Rightarrow d = \frac{h}{3}

The pole (height 20 m) subtends an angle

3 0^{\circ}

at the top of the tower. Using the geometry:

tan 3 0^{\circ} = \frac{20}{d} = \frac{20}{h / 3} = \frac{20 3}{h}

But also, from the triangle at the top of the tower:

tan 3 0^{\circ} = \frac{1}{3} = \frac{h - 20}{d}

Substituting

d = \frac{h}{3}

\frac{1}{3} = \frac{h - 20}{h / 3} = \frac{3 ( h - 20 )}{h}

Solving:

h = 3 (h - 20) \Rightarrow h = 3 h - 60 \Rightarrow 2 h = 60 \Rightarrow h = 30

meters.

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option