JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Heights and Distances

The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is

4 5^{\circ}

. Let R be a point on AQ and from a point B, vertically above R, the angle of elevation of P is

6 0^{\circ}

. If

∠ B A Q = 3 0^{\circ}

A B = d

and the area of the trapezium PQRB is

α

, then the ordered pair

(d, α)

is :

Answer: (A)

Solution

From the given conditions:

Q A = 10

R A = d cos 3 0^{\circ} = \frac{3 d}{2}

QR = 10 - \frac{3 d}{2}

B R = d sin 3 0^{\circ} = \frac{d}{2}

. Using

tan 6 0^{\circ} = \frac{P Q - B R}{QR} = \frac{10 - d /2}{10 - 3 d /2}

, we get

3 = \frac{20 - d}{20 - 3 d}

, which gives

d = 10 (3 - 1)

. The area of trapezium PQRB is

α = \frac{1}{2} (P Q + B R) \cdot QR = \frac{1}{2} (10 + 53 - 5) (10 - 15 + 53) = \frac{1}{2} (75 - 25) = 25

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option