Question

In figure, AB is a chord of the circle and AOC is its diameter such that $\angle ACB={50}^{\circ }$. If AT is the tangent to the circle at the point A, then $\angle$ BAT is equal to

Answer 1

In figure AOC is a diameter of the circle. We know that , diameter subtends an angle ${90}^{\circ }$ at the circle.
$So,\angle ABC={90}^{\circ }In\Delta ABC,\angle A+\angle B+\angle C={180}^{\circ }[Since,sumofallanglesofatriangleis{180}^{\circ }]\Rightarrow \angle A+{90}^{\circ }+{50}^{\circ }={180}^{\circ }\Rightarrow \angle A+140=180\Rightarrow \angle A={180}^{\circ }-{140}^{\circ }={40}^{\circ }\angle Aor\angle OAB={40}^{\circ }$
Now, AT is the tangent to the circle at point A. So , OA is perpendicular to AT.
$\therefore \angle OAT={90}^{\circ }\Rightarrow \angle OAB+\angle BAT={90}^{\circ }Onputting\angle OAB={40}^{\circ },weget\Rightarrow \angle BAT={90}^{\circ }-{40}^{\circ }={50}^{\circ }Hence,thevalueof\angle BATis{50}^{\circ }.$