JEE Main · 2017 · 02 Apr 2017 · Offline · Heights and Distances

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of

A B

and

P

be a point on the ground such that

AP = 2 AB

. If

∠ BPC = β

, then

tan β

is equal to :

(A) $\frac{4}{9}$
(B) $\frac{6}{7}$
(C) $\frac{1}{4}$
(D) $\frac{2}{9}$ ✓ Correct

Answer: (D)

Solution

Let the tower

A B

have height

h

, so its midpoint

C

sits at height

\frac{h}{2}

and

P

is on the ground with

A P = 2 h

. From

P

the top

B

subtends

tan ∠ A P B = \frac{h}{2 h} = \frac{1}{2}

, while

C

subtends

tan ∠ A P C = \frac{h /2}{2 h} = \frac{1}{4}

. The wanted angle is the difference,

β = ∠ A P B - ∠ A P C

, so

tan β = \frac{\frac{1}{2} - \frac{1}{4}}{1 + \frac{1}{2} \cdot \frac{1}{4}} = \frac{1/4}{9/8} = \frac{2}{9}

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option

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