Question

In given figure, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA​

Answer 1

Answer:

Angle PCA = 60°

Step-by-step explanation:

AB is diameter

So, Angle ACB= 90° [Angle made on the boundary of a semicircle]

In triangle ABC

angleA + angle B + angleC = 180° [ASP]

30° + Angle B + 90° = 180°

AngleB = 180-120 = 60°

Angle CBA = 60°

OB = OC [Radius of the same circle]

Angle OBC = angle OCB = 60° [Angle opposite to equal sides are equal]

AngleOCA + Angle OCB = angleACB

angle OCA + 60°=90°

angle OCA = 90°-60° = 30°

angle OCA = 30°

Angle OCP = 90°[Angle on tangent from the center of the circle]

Angle OCA + Angle ACP = angle OCP

30°+angle ACP = 90°

angle ACP = 60°

or alternative method

AO = OC [Radius of same circle]

Angle OAC = angle OCA = 30°[Angle opposite to equal sides are equal]

Angle OCP = 90° [Angle on tangent from the center of the circle]

Angle OCA + Angle ACP = angle OCP

30°+angle ACP = 90°

angle ACP = 60°

Hope it helped you!

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Answer 2

Answer:

ACP= 60 degree