Question

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If $\angle PCA={110}^0$ , find $\angle CBA$ .Hint : Join C with centre O.

• A. ${70}^o$
• B. ${90}^o$
• C. ${50}^o$
• D. ${30}^o$

Answer 1

The correct option is A ${70}^o$

Let $O$ be the center of the circle.

$A,O,B,P$ all are on the same line and $P$ and $C$ are points on the tangent.

$AB$ is a diameter of a circle.

$\therefore$ $\angle BCA={90}^o$ [ Angle inscribe in a semi-circle. ]

$C$ is the point on the circle where the tangent touches the circle.

$\Rightarrow$ So, $\angle OCP={90}^o.$

$\Rightarrow$ $\angle PCA=\angle PCO+\angle OCA$

$\Rightarrow$ ${110}^o={90}^o+\angle OCA$

$\Rightarrow$ $\angle OCA={20}^o$

In $△AOC,$

$\Rightarrow$ $AO=OC$ [ Radius of a circle. ]

$\Rightarrow$ $\angle OCA=\angle CAO={20}^o$

In $△ABC,$

$\Rightarrow$ $\angle CAB+\angle CBA+\angle BCA={180}^o$

$\Rightarrow$ ${20}^o+\angle CBA+{90}^o={180}^o$

$\Rightarrow$ ${110}^o+\angle CBA={180}^o$

$\therefore$ $\angle CBA={70}^o.$