Abcd is a parrelelogram with angle A = 80 degree . the bisectors of angle B and angle C meet at O. find the measure of angles of triangle BCO
Answers
Step-by-step explanation:
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8th
Maths
Let ∠A be 2x
Let ∠D be 2y
Since AB∥DC , sum of angles on same side of the transversal AD will be 180
o
=>2x+2y=180
o
x+y=90
o
After bisection of angles A and D, we get ∠1=x and ∠2=y as per the given figure.
Now in △AOD, we have
x+y+∠AOD=180
o
90
o
+∠AOD=180
o
=>∠AOD=90
o
Step-by-step explanation:
∠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,
c/2=80/2
∴ ∠OCB = 40° [OC is the angle bisector of ∠C]
∠B = 100°
Halving on both sides we get,
b/2=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°
∠OCB = 40°∠OBC = 50°
∠OCB = 40°∠OBC = 50°∠BOC = 90°
hope it's helpful
