Let $A B C$ be an isosceles triangle with $A B = A C$ . Suppose that the angle bisector of angle $B$ meets the side $A C$ at a point $D$ such that $B C = B D + A D$ . Measure of the angle $A$ in degrees, is

100.00

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100.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

100

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

$B C = B D + A D$ (Given)
Let $B C = a, B D = p, A D = q$
$a = p + q$
or $\frac{a}{p} = 1 + \frac{q}{p}$
Using sine law in $△ B D C$ and in $△ A B D,$ we get
$\frac{sin 3 x}{sin 2 x} = 1 + \frac{sin x}{sin 4 x} \Rightarrow \frac{sin 3 x \cdot sin 4 x - sin x \cdot sin 2 x}{sin 2 x \cdot sin 4 x} = 1$
$\Rightarrow (cos x - cos 7 x) - (cos x - cos 3 x) = cos 2 x - cos 6 x$
$\Rightarrow cos 3 x - cos 7 x = cos 2 x - cos 6 x$
$\Rightarrow 2 sin 5 x \cdot sin 2 x = 2 sin 4 x \cdot sin 2 x$
As $sin 2 x \neq 0,$
$sin 5 x = sin 4 x$
$\Rightarrow 5 x + 4 x = 180^{\circ} \Rightarrow x = 20^{\circ}$
Hence, $∠ A = 180^{\circ} - 4 x = 180^{\circ} - 80^{\circ} = 100^{\circ}$

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