An isosceles triangle ABC is right-angled at B. D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the side AB and AC respectively of ΔABC. If AP = a cm, AQ = b cm and ∠BAD = 15°, sin 75° = ?

A. $$\frac{{2b}}{{\sqrt 3 a}}$$

B. $$\frac{a}{{2b}}$$

C. $$\frac{{\sqrt 3 a}}{{2b}}$$

D. $$\frac{{2a}}{{\sqrt 3 b}}$$

Answer: Option C

Solution(By Examveda Team)

According to question from ΔAQD,

∠A = $$\frac{{180 - 90}}{2}$$
∠A = 45°
∠DAQ = 30°
sin 60° = $$\frac{{AQ}}{{AD}}$$
$$\frac{{\sqrt 3 }}{2}$$ = $$\frac{b}{{AD}}$$
AD = $$\frac{{2b}}{{\sqrt 3 }}$$
From ΔAPD
$$\eqalign{ & \sin {75^ \circ } = \frac{{AP}}{{AD}} \cr & \sin {75^ \circ } = \frac{a}{{2b}} \times \sqrt 3 \cr & \sin {75^ \circ } = \frac{{\sqrt 3 a}}{{2b}} \cr} $$