Question

to a circle with centre 0. Jf AB is a diameter and
TIC, ule value of a is 1.
3. In figure, PQ is a tangent at a point
CAB = 30°, find ZPCA.
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Answer 1

Step-by-step explanation:

I hope this may help you

Answer 2

Answer:

Step-by-step explanation:

In triangle ACO

OA = 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

angle PCA = angle PCO - angle CAO

Therefore,

Angle PCA =90°- 30°= 60 ....

I HOPE IT HELPS :):)