Question

in given fig. O is the center of the circle, PQ is a tangent to the circle at A. if angle PAB=58°. find angle ABQ &AQB.​

Answer 1

i hope it hepls..... :)

Answer 2

Answer: angle ABQ is 32 degrees and angle AQB is 26 degrees

Step-by-step explanation:

Joining OA(radius),

OAB + BAP = 90 (radius is perpendicular to the tangent)

=> OAB + 58 = 90

=> OAB = 32

But,

OAB = ABQ (angles in an isosceles triangle are equal as OA and OB are radii)

Therefore, ABQ = 32

Also, BAP = ABQ + AQB (exterior angle property of a triangle)

=> 58 = 32 + AQB

=> AQB = 26

Hence, solved.

Hope this helps ! :)