Question : 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 perpendicular drawn from D on the sides AB and AC respectively of ABC. If $AP = a$ cm , $AQ = b$ cm and $\angle$BAD = 15°, $\sin$ 75°= ?
Option 1: $\frac{2b}{\sqrt 3 a }$
Option 2: $\frac{a}{2b}$
Option 3: $\frac{\sqrt 3a}{2b}$
Option 4: $\frac{2a}{\sqrt 3b}$
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Correct Answer: $\frac{\sqrt 3a}{2b}$
Solution :
Since $\triangle$ABC is a right-angled isosceles triangle,
$\angle$A = $\frac{180 - 90}{2}$ = 45°
$\angle$DAQ = $\angle$A–$\angle$BAD = 45° – 15° = 30°
$\sin$ $\angle$ADQ = $\sin$ 60° = $\frac{AQ}{AD}$
$\frac{b}{AD}$ = $\frac{\sqrt3}{2}$
AD = $\frac{2b}{\sqrt3}$
From $\triangle$APD,
$\sin$ 75° = $\frac{AP}{AD}$ = $\frac{a}{\frac{2b}{\sqrt3}}$ = $\frac{\sqrt3a}{2b}$
Hence, the correct answer is $\frac{\sqrt3a}{2b}$.
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