In the figure, the tangent at a point $ C $ of a circle and a diameter $ A B $ when extended intersect at $ P $. If $ \angle P C A=110^{\circ} $, find $ \angle C B A. $
"

Given:

In the figure, the tangent at a point \( C \) of a circle and a diameter \( A B \) when extended intersect at \( P \).

\( \angle P C A=110^{\circ} \).

To do:
We have to find find \( \angle C B A. \)
 Solution:

$AB$ is a diameter of the circle and from $C$ a tangent is drawn which meets the extended diameter at $P$.

Join $OC$.

Tangent at any point of a circle is perpendicular to the radius through point of contact circle.

This implies,

$OC\ \perp\ PC$

$\angle PCA = 110^o$

$\angle PCO + \angle OCA = 110^o$

$90^o + \angle OCA = 110^o$

$\angle OCA = 110^o-90^o=20^o$

$OC = OA$    (Radii of the circle)

$\angle OCA = \angle OAC = 20^o$     (Angles opposite to equal sides are equal)

$PC$ is a tangent to the circle.

This implies,

$\angle BCP = \angle CAB = 20^o$     (Angles in alternate segment are equal)

In $\triangle PBC$,

$\angle P + \angle C + \angle A= 180^o$

$\angle P = 180^o - (\angle C + \angle A)$

$\angle P = 180^o - (110^o + 20^o)$

$\angle P = 180^o - 130^o = 50^o$

In $\triangle PBC$,

$\angle BPC + \angle PCB + \angle PBC = 180^o$

$50^o + 20^o + \angle PBC = 180^o$

$\angle PBC = 180^o - 70^o$

$\angle PBC = 110^o$

$APB$ is a straight line.

$\angle PBC + \angle CBA = 180^o$

$\angle CBA = 180^o - 110^o = 70^o$

Therefore, \( \angle C B A=70^o \).

Tutorialspoint

e

Updated on: 10-Oct-2022

32 Views

Kickstart Your Career

Get certified by completing the course

Get Started