Question

In the given PQR is a tangent at a point to a circle with center O. If AB is a diameter and Angle CAB=30 degree , Find angle PCA.

Answer 1

Answer:

∠PCA = 60°

Step-by-step explanation:

Given:- A Circle with centre O and a tangent PQ, AB is a diameter, and ∠CAB = 30°

To find:- ∠PCA

Construction:- Join O to C then OC becomes the radius of the Circle.

Proof:-

We know that,

∠PCO = 90°

[Radius is perpendicular to the tangent at the point of contact]

also,

∠PCA + ∠ACO = ∠PCO = 90°

∠PCA = 90 - ∠ACO ----- 1

We are given that,

∠CAB = 30°

In ΔAOC,

OA = OC [Radii of the same Circle]

Thus,

∠CAB = ∠ACO [Properties of Isocseles Triangles]

hence, ∠ACO = 30° ------ 2

then, from eq.1 and eq.2 we get,

∠PCA = 90 - 30

∠PCA = 60°

Hope it helped and you understood it........All the best