Question

in the figure AB is the diameter with center 0 , AP is a tangent . If OCB = 25° find the APB
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Answer 1

angle OCB=angleOBC

as because oc and ob are radius of same circle

now in ∆ABP

angle ABP=25°

angle PAB=90°

so now angle APB=180°-(90°+25°)

=180°-115

= 65°

so the answer is 65°

Answer 2

Answer:

The measure of ∠APB is 65°

Step-by-step explanation:

Given the figure, AB is the diameter with center O , AP is a tangent. If ∠ OCB = 25°.

we have to find the ∠APB.

As, OC=OB  (∵Both are radii of same circle)

⇒ ∠OCB=∠OBC=25°

Now in ∆ABP

∠ABP=25°

∠PAB=90°   (∵ radius from center is perpendicular on the tangent)

By angle sum property of triangle

∠ABP+∠BAP+∠APB=180°

⇒ ∠APB=180°-(90°+25°)=180°-115  = 65°

Hence,  ∠APB is 65°