ABCD is a parallelogram with ∠A= 80°. The internal bisectors ∠B a

1.	ABCD is a parallelogram with ∠A= 80°. The internal bisectors ∠B and ∠C meet at O. Find the measure of three angles of triangle BCO
Answer» Step-by-step EXPLANATION: ∠A = 80° We know that the opposite ANGLES of a parallelogram are equal. ∠A = ∠C = 80° And ∠OCB = (1/2) × ∠C = (1/2) × 80° = 40° ∠B = 180° – ∠A (the sum of interior angles on the same side of the TRANSVERSAL is 180) = 180° – 80° = 100° Also, ∠CBO = (1/2) × ∠B = (1/2) × 100° = 50° By the angle sum property of TRIANGLE BCO, ∠BOC + ∠OBC + ∠CBO = 180° ∠BOC = 180° – (∠OBC + CBO) = 180° – (40° + 50°) = 180° – 90° = 90° Hence, the measure of all the three angles of a triangle BCO is 40°, 50° and 90°.

ABCD is a parallelogram with ∠A= 80°. The internal bisectors ∠B and ∠C meet at O. Find the measure of three angles of triangle BCO

Answer»

Step-by-step EXPLANATION:

∠A = 80°

We know that the opposite ANGLES of a parallelogram are equal.

∠A = ∠C = 80°

And

∠OCB = (1/2) × ∠C

= (1/2) × 80°

= 40°

∠B = 180° – ∠A (the sum of interior angles on the same side of the TRANSVERSAL is 180)

= 180° – 80°

= 100°

Also,

∠CBO = (1/2) × ∠B

= (1/2) × 100°

= 50°

By the angle sum property of TRIANGLE BCO,

∠BOC + ∠OBC + ∠CBO = 180°

∠BOC = 180° – (∠OBC + CBO)

= 180° – (40° + 50°)

= 180° – 90°

= 90°

Hence, the measure of all the three angles of a triangle BCO is 40°, 50° and 90°.