Question

In the given figure, O is the centre of a circle, AOC is its diameter such that $\angle$ ACB = ${50}^o$ . If AT is the tangent to the circle at the point A then $\angle$ BAT = ?

(a) ${40}^o$ (b) ${50}^o$ (c) ${60}^o$ (d) ${65}^o$

Answer 1

In the figure ∠CBA =90°(as angle in a semicircle is a right angle)
And ∠CAT = 90° (as radius to the point of tangency is always perpendicular to the tangent line)
Now in ∆ABC ,
∠CAB+∠ABC+∠ACB = 180°(sum of the interior angles of a triangle)
⇒∠CAB +90°+50°=180°
⇒∠CAB =180°−140° =40°
Now ∠CAB +∠BAT =∠CAT ⇒∠BAT =∠CAT −∠CAB ⇒∠BAT =90°−40° =50°

(b) 50o