Question

In the given figure, ΔOQP = ΔOAB, angle OPQ = 52° and angle BOQ = 104°. Find angle OAB

Answer 1

Answer:

Step-by-step explanation:

PQ is parallel to AB

So, Angle OPQ = Angle OBA (because, alternate interior angles)

Angle OBA = 52 Degrees

Angle QOB = Angle POA = 104 Degrees ( Vertically Opposite Angles) - (1)

Angle POQ = Angle AOB ( Vertically Opposite Angles) - (2)

Angle(QOB + POA + POQ + AOB) = 360 Degrees

Angle(2QOB + 2AOB) = 360 Degrees [ from (1) and (2)]

208 + 2 AOB = 360

2 AOB = 360 - 208 = 152

AOB = 152/2 = 76 Degrees

Therefore, Angle AOB + Angle OBA + Angle OAB = 180 Degrees

76 + 52 + OAB = 180

128 + AOB = 180

AOB = 180 - 128 = 52

Angle AOB = 52 Degrees

Answer 2

∠QOP + ∠BOQ = 180° (Linear pair)

∠QOP + 104° = 180°

∠QOP = 180° – 104° = 76⁰

∠OPQ + ∠QOP + ∠PQO = 180° (Angles of a triangle) 52° + 76° + ∠PQO = 180⁰

128° + ∠PQO = 180⁰

∠PQO = 180⁰ - 128⁰ = 52°

∠OAB = ∠PQO = 52° (vertically opposite angles)