Question

O is the center of the circle. If $\angle BAC={50}^{\circ }$, find $\angle OBC$

Answer 1

Given a circle with it's center at O

$\angle BAC={50}^{\circ }$

$\angle BOC=2\angle BAC$ (Angle made by a chord at centre is twice of angle it makes with circumference.)

$=2\left({50}^{\circ }\right)$

$={100}^{\circ }$

In $△OBC,OB=OC=\text{radius}$

$\Rightarrow \angle OBC=\angle OCB$ ( opposite angles of equal sides of a $△$ )

Sum of angles in a triangle = ${180}^{\circ }$

$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }....\left(1\right)$

Let$\angle OBC=\angle OCB=x$

$⟹x+x+\angle BOC={180}^{\circ }$ From $\left(1\right)$

$2x={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

$⟹x={40}^{\circ }$

$\therefore$$\angle OBC=x={40}^{\circ }$