Multiple choice

If the median AD of a triangle ABC divides the angle $\displaystyle \angle BAC$ in the ratio 1 :2, then $\dfrac{\sin B}{\sin C}$ is equal to

  1. $2 \cos \dfrac{A}{3}$
  2. $\dfrac12 \: \sec \dfrac A3$
  3. $\dfrac12 \: \sin \dfrac A3$
  4. $2 \: \mathrm{cosec} \dfrac A3 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the sine rule in triangles ABD and ACD, and the fact that AD bisects angle A into A/3 and 2A/3, we can relate the sides and angles. The ratio sin(B)/sin(C) simplifies to (AC/AB) * (sin(angle ADB)/sin(angle ADC)). Applying the angle bisector theorem and sine rule properties leads to the result 1/2 sec(A/3).

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