Question

In ΔABC, bisectors of ∠A and ∠B intersect at point O. If ∠C=70°. Find measure of ∠AOB.

Answer 1

Answer:

∠AOB = 125°

Step-by-step explanation:

Let say ∠A = x°  & ∠ B = y°

∠A + ∠ B + ∠C = 180°  ( Sum of angles of Δ ABC)

=> x° + y° + 70° = 180°

=> x° + y°  = 110°

bisectors of ∠A and ∠B intersect at point O

=> ∠BAO =  ∠A/2 = (x/2)°

& ∠ABO = ∠B/2 = (y/2)°

in ΔAOB

∠BAO + ∠ABO + ∠AOB = 180°

=> (x/2)° + (y/2)° + ∠AOB = 180°

=> x° + y° + 2∠AOB = 360°

=> 110° + 2∠AOB = 360°

=> 2∠AOB = 250°

=> ∠AOB = 125°

measure of ∠AOB = 125°

Answer 2

Answer:

∠AOB = 125°

Step-by-step explanation:

In the triangle ABC,

∠C = 70°

So,

Let us say,

∠A = x and ∠B = y

So,

∠A + ∠B +∠C = 180° (∵ Sum of internal angles of a triangle is 180°.)

∠A + ∠B + 70° = 180°

x + y = 110° ...........(1)

Now,

In triangle AOB,

∠BAO + ∠ABO + ∠AOB = 180° (∵ Sum of internal angles of a triangle is 180°.)

$\frac{x}{2}+\frac{y}{2}+AOB=180\\AOB=180-(\frac{x+y}{2})$

(Because, AO and BO are the bisectors of the angles A and B.)

Now,

On putting the value from the equation (1) we get,

$AOB=180-\frac{110}{2}=180-55=125$

Therefore,

∠AOB = 125°