Question

plz solve in detail.
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Answer 1

Thanks for asking the question!

ANSWER::

Given=>

DC is a tangent at C to the circle with centre O

∠ACD = 42°

OC= OA ( Radii of same circle)

∠OCA = ∠OAC = 90° - 42° = 48°

In ΔDAC , ∠DAC = 92° ( By Angle Sum Property of Triangle)

∠DAC+ ∠CAO + ∠OAB = 180°

92°+42°+∠OAB = 180°

∠OAB = 180–134 = 46°

OA=OB ( Radii of circle)

∠OAB = ∠OBA = 46°

In ΔOAB

∠AOB= 180°- (2x46°)= 180°–92°= 88°

Hope it helps!

Answer 2

Answer:-

∠ACD = 42°

OC= OA

∠OCA equal to ∠OAC and this equal to 90° - 42° = 48°

By ASP(By Angle Sum Property of the Triangle)

∠ DAC and ∠ DAC = 92°

∠ DAC+ ∠ CAO + ∠ OAB = 180°

92°+42°+ ∠ OAB = 180°

∠ OAB = 180–134 = 46°

OA=OB

∠ OAB = ∠ OBA = 46°

[[[[[ΔOAB]]]]]

∠ AOB= 180°- (2x46°)= 180°–92°= 88°