Math, asked by dineshpayasidgs, 1 year ago

In the figure AB is the diameter of the circle and PQ is tangent at B , given that

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Answered by tuka81
0

In Δ ACO,

OA=OC (Radii of the same circle)

Therefore,

ΔACO is an isosceles triangle.

∠CAB = 30° (Given)

∠CAO = ∠ACO = 30° (angles opposite to equal sides of an isosceles triangle are equal)

∠PCO = 90° …(radius is drawn at the point of

contact is perpendicular to the tangent)

Now ∠PCA = ∠PCO – ∠CAO

Therefore,

∠PCA = 90° – 30° = 60°


lovekush65: angle P and C =70° jha tk
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