Question

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

• A. $(\frac{-\sqrt{3}a}{2},\frac{a}{2})$
• B. $(\frac{-\sqrt{3}a}{2},\frac{a}{4})$
• C. $(\frac{a}{2},\frac{\sqrt{3}a}{2})$
• D. $(\frac{-\sqrt{3}a}{2},\frac{-a}{2})$

Answer 1

Answer: (A), (D)

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

Since $\angle B=\angle C={75}^{\circ }$, we have
$\angle BAC={30}^{\circ }$
or $\angle BOC={60}^{\circ }$
$\therefore B\equiv (-a\cos {30}^{\circ },a\sin {30}^{\circ })$
$\equiv (\frac{-\sqrt{3}a}{2},\frac{a}{2})$
and $C\equiv (-\frac{\sqrt{3}a}{2},\frac{-a}{2})$