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1995

JEE Main 1995 Mathematics question on Trigonometry

In a △ABC\triangle ABC, ∠B=π3\angle B = \dfrac{\pi}{3} and ∠C=π4\angle C = \dfrac{\pi}{4}. Let DD divides BCBC internally in the ratio 1:31:3, then sin⁡∠BADsin⁡∠CAD\dfrac{\sin \angle BAD}{\sin \angle CAD} is equal to
  1. A.
    16\dfrac{1}{\sqrt{6}}
  2. B.
    13\dfrac{1}{3}
  3. C.
    13\dfrac{1}{\sqrt{3}}
  4. D.
    23\sqrt{\dfrac{2}{3}}
Show answer and solution

Correct option: A

Solution

Using the generalized rule: in triangle ABC with D on BC dividing it as BD:DC = 1:3, apply the sine rule in triangles ABD and ACD.

In △ABD: BD/sin∠BAD = AD/sin B, so sin∠BAD = BD·sinB/AD
In △ACD: DC/sin∠CAD = AD/sin C, so sin∠CAD = DC·sinC/AD

Therefore: sin∠BAD/sin∠CAD = (BD·sinB)/(DC·sinC) = (1/3)·(sinB/sinC)

Since ∠B = π/3 and ∠C = π/4:
sin(π/3)/sin(π/4) = (√3/2)/(1/√2) = (√3/2)·(√2) = √6/2

So sin∠BAD/sin∠CAD = (1/3)·(√6/2) = √6/6 = 1/√6.

[1995]

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