Question

In the figure, $\angle BAE={60}^{\circ }$ and $\angle CAF={80}^{\circ }$, then $\angle BDA$ is

• A. ${30}^{\circ }$
• B. ${80}^{\circ }$
• C. ${60}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Let "O" be the centre of the circle and OA, OB are the radii of the circle

Then $\angle OAE={90}^{\circ }$, Because tangent on a circle is perpendicular to its radius.
$\Rightarrow \angle OAB={90}^{\circ }-{60}^{\circ }={30}^{\circ }$
$\angle OBA=\angle OAB$ (Since OA and OB are equal.)
Then $\angle BOA={180}^{\circ }-({30}^{\circ }+{30}^{\circ })={180}^{\circ }-{60}^{\circ }={120}^{\circ }$ (Sum angle in the triangles is 180)
Then $\angle BDA=\frac{120}{2}={60}^{\circ }$ (The angle at the centre is twice the angle at the circumference.)