Question

An isosceles $△ABC$ is inscribed in a circle $x^2+y^2=a^2$ with the vertex $A$ at $(a,0)$ and the base angles $B$ and $C$ each equal to ${75}^0$ then length of the base $BC$ is

• A. $\frac{a}{2}$
• B. $a$
• C. $\frac{2a}{\sqrt{3}}$
• D. $\frac{\sqrt{3}a}{2}$

Answer 1

Answer: (B)

$a$

$\angle B=\angle C={75}^0$

$\Rightarrow \angle BAC={30}^0\Rightarrow \angle BOC={60}^0$

$\Rightarrow BOC$ is an equilateral triangle

$\Rightarrow BC=OB=$ the radius of the circle

$\Rightarrow BC=a$