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) In the given figure, ABCD is a parallelogram with∠ A = 80o .The bisectors of ∠ B and ∠ C meet each other at O. Find the measure of three angles of ∆BCO.

) In the given figure, ABCD is a parallelogram with∠ A = 80o
.The bisectors of ∠ B and
∠ C meet each other at O. Find the measure of three angles of ∆BCO. 

Grade:8

1 Answers

Aman Tomar
17 Points
2 months 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°

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