Question

In the given figure, PAQ is the tangent. BC is the diameter of the circle. If $\angle BAQ={60}^{\circ }$, then $\angle ABC$ = _____.

• A. ${25}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B

${30}^{\circ }$

Join OA

As the tangent at any point of a circle is perpendicular to the radius through the point of contact

$\angle OAQ={90}^{\circ }$

$\angle OAB=\angle OAQ-\angle BAQ$

$\angle OAB={90}^{\circ }–{60}^{\circ }$

$\angle OAB={30}^{\circ }$

$OA=OB$ ( $\because$ radius of the same circle)

$\Rightarrow △OAB$ is an isosceles triangle.

$\angle OAB=\angle OBA$ ( $\because$ base angles of an isosceles triangle)

Therefore $\angle OBA=\angle OAB={30}^{\circ }$

$\angle ABC=\angle OBA={30}^{\circ }$