Question

In the adjoining figure, PT is a tangent at point C of the circle. O is the circumcentre of $\Delta ABC$. If $\angle ACP={118}^0$, then the measure of $\angle x.$ is

• A. ${28}^0$
• B. ${32}^0$
• C. ${42}^0$
• D. ${38}^0$

Answer 1

Answer: (A)

${28}^0$

Join OC.
OC is the radius and Pt is the tangent to circle at point C
$\therefore OC\perp PT$
$\therefore \angle OCP={90}^0$
given $\angle ACP={118}^0$
$\therefore \angle ACO=\angle ACP-\angle OCP$
$={118}^0={90}^0$
$\angle ACO={28}^0$
Since O is the circumcentre, thus $OA=OC$ (radius)
$\therefore \angle OAC=\angle OCA$
$\therefore x={28}^0$