Aman Tomar
Last Activity: 3 Years ago
∠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}
2
C
= \sf\frac{80}{2}
2
80
∴ ∠OCB = 40° [OC is the angle bisector of ∠C]
∠B = 100°
Halving on both sides we get,
\sf\frac{B}{2}
2
B
= \sf\frac{100}{2}
2
100
∴ ∠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°
