Question

An isoscels triangle is inscribed in the circle $x^2+y^2=a^2$. If the vertex of the triangle is at $A(a,0)$ and the base angles $B$ and $C$ be each ${75}^{\circ }$, then the coordinates of $B$ and $C$ can be

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

Answer 1

Answer: (B), (D)

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

Given equation of circle is
$x^2+y^2=a^2$
Center is $(0,0)$ and radius $a$
Given $ABC$ is an isosceles triangle with $\angle B={75}^0,\angle C={75}^0$
$\Rightarrow \angle A={30}^0$
$\Rightarrow \angle BOC={60}^0$

(Angle subtended at the center of circle is twice the angle subtended at any point on the circle)
Now consider $\Delta OBD$ and $\Delta OCD$
$OB=OC$ (radius)
$BD=DC$ (Perepndicular from center of circle to the chord bisects the chord)
$OD=OD$ (common)
$\Delta OBD\cong \Delta OCD$
$\Rightarrow \angle OBD=\angle OCD$
So, $\angle OBD+\angle OCD+{60}^0={180}^0$
$\Rightarrow \angle OBD=\angle OCD={60}^0$
Hence, $\Delta BOC$ is equilateral
$\Rightarrow BC=OB=OC=a$
$\Rightarrow BD=DC=\frac{a}{2}$
In right triangle OBD,
$\sin {60}^0=\frac{OD}{OB}$

$\Rightarrow OD=\frac{a\sqrt{3}}{2}$
Clearly, the coordinates of $B$ and $C$ are $(-\frac{a\sqrt{3}}{2},\frac{a}{2})$ and $(-\frac{a\sqrt{3}}{2},-\frac{a}{2})$

Hence, option B and D.