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° = ?

Answer: Option C

Solution

Answer: Option C

Solution:

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
$$\begin{array}{rl}& \sin 75^{\circ }=\frac{AP}{AD}\\ & \sin 75^{\circ }=\frac{a}{2b}\times \sqrt{3}\\ & \sin 75^{\circ }=\frac{\sqrt{3}a}{2b}\end{array}$$