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JEE Main 2017

JEE Main 2017 Mathematics question on Trigonometry

Let a vertical tower ABAB have its end AA on the level ground. Let CC be the mid-point of ABAB and PP be a point on the ground such that AP=2ABAP = 2AB. If ∠BPC=β\angle BPC = \beta, then tan⁡β\tan\beta is equal to
  1. A.
    67\dfrac{6}{7}
  2. B.
    14\dfrac{1}{4}
  3. C.
    29\dfrac{2}{9}
  4. D.
    49\dfrac{4}{9}
Show answer and solution

Correct option: C

Solution

Let AB = 2h, so AC = CB = h, and AP = 2AB = 4h.

tan(∠BPA) = 2h/4h = 1/2, and tan(∠CPA) = h/4h = 1/4.

Since β = ∠BPC = ∠BPA - ∠CPA, use the subtraction formula:

tan β = (1/2 - 1/4)/(1 + 1/2·1/4) = (1/4)/(9/8) = 2/9.

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