Question

An isosceles triangles $ABC$ is inscribed in a circle $x^2+y^2=a^2$ with the vertex A at $(a,0)$ and the base angle $B$ and $C$ each equal to 75${}^{\circ }$, then coordinates of an end point of the base are

• A. $(\frac{-\sqrt{3}a}{2},\frac{a}{2})$
• B. $(-\frac{\sqrt{3}a}{2},a)$
• C. $\left(\frac{a}{2},\frac{\sqrt{3}a}{2}\right)$
• D. $\left(\frac{\sqrt{3}a}{2},-\frac{a}{2}\right)$

Answer 1

Answer: (A)

$(\frac{-\sqrt{3}a}{2},\frac{a}{2})$

Since, $\angle B=\angle C={75}^o$ for triangle $BAC$
$\Rightarrow$ $\angle BAC={30}^o$
$\Rightarrow$ $\angle BOC={60}^o$
$\Rightarrow$ B has coordinates $(-a\cos {30}^o,a\sin {30}^o)$
or $(\frac{-\sqrt{3}a}{2},\frac{a}{2})$ and those of C are $(-\frac{\sqrt{3}a}{2},-\frac{a}{2})$