Question

In 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

Step-by-step explanation:

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In triangle ACO

In triangle ACOOA = OC (Radii of the same circle) Therefore,

In triangle ACOOA = OC (Radii of the same circle) Therefore,Triangle ACO is an isosceles triangle.

In triangle ACOOA = OC (Radii of the same circle) Therefore,Triangle ACO is an isosceles triangle.angle CAB = 30° (Given)

In triangle ACOOA = OC (Radii of the same circle) Therefore,Triangle ACO is an isosceles triangle.angle CAB = 30° (Given)angle CAO = ACO 30° (angles opposite to equal sides of an isosceles triangle are equal)

In triangle ACOOA = OC (Radii of the same circle) Therefore,Triangle ACO is an isosceles triangle.angle CAB = 30° (Given)angle CAO = ACO 30° (angles opposite to equal sides of an isosceles triangle are equal)angle PCO=90(radius is drawn at the point of contact is perpendicular to the tangent)

In triangle ACOOA = OC (Radii of the same circle) Therefore,Triangle ACO is an isosceles triangle.angle CAB = 30° (Given)angle CAO = ACO 30° (angles opposite to equal sides of an isosceles triangle are equal)angle PCO=90(radius is drawn at the point of contact is perpendicular to the tangent)Now

In triangle ACOOA = OC (Radii of the same circle) Therefore,Triangle ACO is an isosceles triangle.angle CAB = 30° (Given)angle CAO = ACO 30° (angles opposite to equal sides of an isosceles triangle are equal)angle PCO=90(radius is drawn at the point of contact is perpendicular to the tangent)Nowangle PCA = angle PCO - angle CAO

In triangle ACOOA = OC (Radii of the same circle) Therefore,Triangle ACO is an isosceles triangle.angle CAB = 30° (Given)angle CAO = ACO 30° (angles opposite to equal sides of an isosceles triangle are equal)angle PCO=90(radius is drawn at the point of contact is perpendicular to the tangent)Nowangle PCA = angle PCO - angle CAOTherefore,

In triangle ACOOA = OC (Radii of the same circle) Therefore,Triangle ACO is an isosceles triangle.angle CAB = 30° (Given)angle CAO = ACO 30° (angles opposite to equal sides of an isosceles triangle are equal)angle PCO=90(radius is drawn at the point of contact is perpendicular to the tangent)Nowangle PCA = angle PCO - angle CAOTherefore,Angle PCA =90°- 30°=60°

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Answer 2

Step-by-step explanation:

Answer -

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