Question

In an isosceles triangle the sine of the base angle is three times as large as the cosine of the vertex angle. Find the sine of the base angle.

Answer 1

Let the vertex angle be $x$ and the base angles be $y$.

we know that, in a triangle, sum of all the angles is 180

Thus $x+2y=180⟹x=180-2y$

Also given $\sin y=3\cos x$

$⟹\sin y=3\cos (180-2y)=-3\cos 2y$

$⟹3\cos 2y+\sin y=0$

$⟹3(1-2{\sin }^{2}y)+\sin y=0$

Assume $\sin y=a$

$⟹6a^2-a-3=0$

$⟹a=\frac{1\pm \sqrt{73}}{12}$

since $0\le \sin y\le 1$ for the base angle of an isosceles triangle, we discard the negative value and consider only the positive value.

Thus $\sin y=\frac{1+\sqrt{73}}{12}$