Question

In triangle ABC ,bisectors of angle A and angle B intersect at point O.If angle C=70.find measure of angle AOB.​

Answer 1

Answer:

∠AOB=125°

Step-by-step explanation:

In the Δ ABC :

Given: ∠ C=70°

Bisector of ∠ A and ∠ B are intersect each other at point O.

To find: ∠AOB=?

Solution:

As we know that ,In any triangle:

Sum of all angles=180° [ Angle sum property of triangles]

So in ΔABC, ∠A+∠B+∠C=180°....................(1)[From figure]

∠A=70°  (GIVEN)

Put the value of ∠A in equation(1)

70°+∠B+∠C=180°

∠B+∠C=110°

∠B/2+∠C/2=55°

∠OAB+∠OBA=55°...............(2) [Because bisectors of ∠B and ∠C are intersecting  at point O.]

In ΔAOB :

∠OAB+∠OBA+∠AOB=180° [Angle sum property of triangle]..........(3)

Put the value of equation(2) in equation(3), we get:

55°+∠AOB=180°

∠AOB=180°-55°

∠AOB=125°

Answer 2

Step-by-step explanation:

In a triangle,

the sum of interior angle 180°

Thus, in △ABC ,

∠A+∠B+∠C=180 ∘

∠C=70 ∘

Hence,

∠A+∠B=130 ∘

Bisector of ∠A and∠B intersect at point O then,

2∠A + 2∠B

= 65 ∘

∠OAB+∠OBA=65∘

Now, in △OAB, t

∠OAB+∠OBA+∠AOB=180∘