Question

Question 12
The tangent at a point C of a circle and a diameter AB when extended intersect at P. If $\angle PCA={110}^{\circ }$ , find $\angle$ CBA.

Answer 1

Here, AB is a diameter of the circle from point C and a tangent is drawn which meets at a point P.

Join OC. Here OC is radius
Since tangent at any point of a circle is perpendicular to the radius through point of contact circle.
$\therefore$ OC $\perp$ PC
Now $\angle$ PCA = ${110}^{\circ }$ [given]
$\Rightarrow \angle PCO+\angle OCA={110}^{\circ }\Rightarrow {90}^{\circ }+\angle OCA={110}^{\circ }\Rightarrow \angle OCA={20}^{\circ }$
$\because$ OC= OA = Radius of circle
$\angle$ OCA = $\angle$ OAC = ${20}^{\circ }$
[Since two sides are equal, then their opposite angles are equal]
Since PC is a tangent So,$\angle BCP=\angle CAB={20}^{\circ }$
[ angle in a alternate segment are equal]
In $\Delta$ ABC, $\angle P+\angle C+\angle A={180}^{\circ }$
$\angle P={180}^{\circ }-(\angle C+\angle A)$
$={180}^{\circ }=({110}^{\circ }+{20}^{\circ })$
$={180}^{\circ }-{130}^{\circ }={50}^{\circ }$
In $\Delta$ PBC,
$\angle BPC+\angle PCB+\angle PBC={180}^{\circ }$
[ Sum of all interior angles of any triangle is ${180}^{\circ }$]
$\Rightarrow {50}^{\circ }+{20}^{\circ }+\angle PBC={180}^{\circ }$
$\Rightarrow \angle PBC={180}^{\circ }-{70}^{\circ }$
$\Rightarrow \angle PBC={110}^{\circ }$
Since, APB is a straight line
$\therefore \angle PBC+\angle CBA={180}^{\circ }$
$\Rightarrow \angle CBA={180}^{\circ }-{110}^{\circ }={70}^{\circ }$