Question

4. In the following figure. PQ is a tangent at a point to circle with centre 0. If AB is a
diameter and 2C AB= 30°, then find ZPCA.
300 A
E​

Answer 1

Step-by-step explanation:

Given: PQ is a tangent. AB is a diameter, ∠CAB=30

o

To find: ∠PCA=?

In ΔAOC,

∠CAB=∠OCA (Angles opposite to equal sides are equal)

So, ∠CAB=30

o

=∠OCA

Since OC⊥PQ (Tangent is perpendicular to the radius at point of contact)

∠PCO=90

o

∠OCA+∠PCA=90

o

30

o

+∠PCA=90

o

∠PCA=90

o

−30

o

Therefore, ∠PCA=60

Answer 2

Step-by-step explanation:

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