1.

ABCD is a parallelogram with angle A equal to 80°. The internal bisectors of Angle B and angle C meet at O. Find measures of the three angles of triangle BCO.

Answer»

Answer:

∠OBC = 50°

∠OCB = 40

∠BCO = 90°

Step-by-step EXPLANATION:

Given.

In ║gms ABC,

∠A = 80°

To Find.

∠OBC

∠OCB

∠BCO

Solution.

In Parallelogram ABCD,

∠A = ∠C [OPPOSITE angles of a ║gm are EQUAL]

∠B = ∠D [Opposite angles of a ║gm are equal]

∴ ∠A = ∠C = 80°

∠A + ∠D = 180° [Co-interior angles]

80° + ∠D = 180°

∠D = 180° - 80°

∠D = 100°

But ∠B = ∠D

∴ ∠B = 100°

∠C = 80°

Halving on both sides we get,

\sf\frac{C}{2} = \sf\frac{80}{2}

∴ ∠OCB = 40° [OC is the ANGLE bisector of ∠C]

∠B = 100°

Halving on both sides we get,

\sf\frac{B}{2} = \sf\frac{100}{2}

∴ ∠OBC  = 50° [OB is the angle bisector of ∠B]

In ΔBOC

∠OBC + ∠OCB + ∠BOC = 180°

50° + 40° + ∠BOC = 180°

90° + ∠BOC = 180°

∠BOC = 180° - 90°

∴ ∠BOC = 90°

Final answers

∠OCB = 40°

∠OBC  = 50°

∠BOC = 90°


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