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

\(\frac{\sqrt {3a}}{2b}\)

∠BAD  = 15°

∠DAC = 45° - 15° = 30°

∠ADX = 90° - 15° = 75°

∠ADY = 90° - 30° = 60°

in ΔADQ,

sin ∠ADQ = sin 60°

\(\frac{\sqrt {3}}{2}=\frac{b}{AD}⇒AD=\frac{2b}{\sqrt {3}}\)

in ΔAXD,

\(sin 75° =\frac{AX}{AD}\)

\(sin 75° =\frac{a}{\frac{2b}{\sqrt {3}}}=\frac{\sqrt {3}a}{2b}\)