Question

In fig 9.4, AB is a chord of the circle and AOC is its diameter such that $\angle ACB={50}^0$. If AT is the tangent to the circle at the point A, then $\angle BAT$ is equal to

• A. ${65}^0$
• B. ${60}^0$
• C. ${50}^0$
• D. ${40}^0$

Answer 1

Answer: (C)

${50}^0$

from the given figure we know a tangent is perpendicular to the circle

so $\angle TAC=\angle TAB+\angle BAC=90$ [equation 1]

now in triangle ABC $\angle ABC=90$ [since the triangle is in a semi circle]

$\angle ABC+\angle BCA+\angle CAB=180$

therefore $\angle CAB=180-90-50=40$

so from equation 1 we get

$\angle TAB=90-\angle BAC=90-40=50$