Question

solve this question very fast

answer with solution​

Answer 1

Answer:

AB=Ac

BAC = 72

angle made by two equal sides are equal

ABC=ACB

let ABC=ACB

in triangle ABC

BAC + ABC + ACB= 108

by using triangle angle sum property

suitable the value then we get

X + 72 + x= 108

2x=108-72 =108

X= 108÷2 =54

ABC +ACB =54

Central angle is twice the inscribed angle

AOB = 2 × ACB =2×54=108

AOB = 108

we know that radius is perpendicular to tangent

OA is perpendicular to P A and o b is perpendicular to PB.

OAP = OBA = 90

in quadrilateral OAPB

OAP + OBP+ APB + AOB = 360

substitute the value then we get

90 + 90 + 108 + APB = 360

APB = 360 -288=72

APB = 72

I hope you satisfied this answer

Answer 2

Answer:

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