JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Heights and Distances

Two poles

A B

of length a meter and

C D

of length

a + b (b \neq = a)

metres are erected at the same horizontal level with bases at

B

and

D

. If

B D = x

and

tan ∠ A C B = \frac{1}{2}

, then:

(A) $x^{2} + 2 (a + 2 b) x - b (a + b) = 0$
(B) $x^{2} - 2 a x + a (a + b) = 0$
(C) $x^{2} + 2 (a + 2 b) x + a (a + b) = 0$
(D) $x^{2} - 2 a x + b (a + b) = 0$ ✓ Correct

Answer: (D)

Solution

Take

B = (0, 0)

D = (x, 0)

A = (0, a)

and

C = (x, a + b)

. At

C

the vectors to

A

and

B

are

C A = (- x, - b)

and

C B = (- x, - (a + b))

, so

tan ∠ A C B = \frac{C A \times C B}{C A \cdot C B} = \frac{a x}{x ^{2} + b ( a + b )} = \frac{1}{2} .

This gives

2 a x = x^{2} + b (a + b)

, i.e.

x^{2} - 2 a x + b (a + b) = 0

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option

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