An isosceles triangle ABC is inscribed in a circle x² + y² = a² with the vertex A at (a, 0) and the base angles B and C each equal to 75⁰ then coordinates of an end point of the base are

Given  an isosceles triangle ABC inscribed in a   circle $x^2+y^2=a^2$

Since $\overline{)\mathrm{B}}=\overline{)\mathrm{C}}={75}^{\circ }$
$\begin{array}{l}\Rightarrow \overline{)\mathrm{BAC}}={30}^{\circ }\Rightarrow \overline{)△\mathrm{BOC}}={60}^{\circ }\\ \Rightarrow \text{ the triangle is an equilateral triangle}\\ \angle \mathrm{AOB}={150}^{\circ };\text{B},\text{C are }\left(\text{ a }\cos {150}^{\circ },\pm \text{ a sin }{150}^{\circ }\right)\\ =(\frac{-\sqrt{3}a}{2},\frac{a}{2}) and (\frac{-\sqrt{3}a}{2},-\frac{a}{2})\end{array}$