Question

21. In Fig., BO and CO bisect angle B and C respectively. If angle BOC = 112°, then angle A = B с
(a) 88° (b) 64° (c) 28° (d) 44°
​

Answer 1

Answer:

Correct option is

A

90

o

−

2

1

∠A

As BO and CO are the angle bisectors of external angles of△ABC, Then

∠1=∠2

∠4=∠3

We know, ∠A+∠ABC+∠ACB=180

∘

…eqn(1)

And ∠ABC=180−2∠1

∠ACB=180−2∠4

Putting it in the eqn (1), we get

∠A+180−2∠1+180−2∠4=180

⇒∠1+∠4=90+

2

1

∠A…eqn(2)

Also we know from the figure, ∠BOC+∠1+∠4=180

∘

∠BOC=180−∠1−∠4

From eqn (2)

∠BOC=180−90−

2

1

∠A

⇒∠BOC=90

∘

−

2

1

∠A

Answer 2

Given: In Fig., BO and CO bisect angle B and C respectively. If angle BOC = 112°.

To find: The angle A.

Solution:

In the triangle BOC, BO and CO are equal in length and hence, the angles ∠OBC and ∠OCB are also equal. Let each of those two angles be x. Since the sum of all the angles of a triangle is 180, the following equation can be formed.

$112 + x + x = 180$

$2x = 68$

$x = 34$

As BO and CO bisect ∠B and ∠C, they can be written as follows.

$\angle ABC = \angle OBC * 2$

           $= 34 * 2$

           $= 68$

$\angle ACB = \angle OCB * 2$

           $= 34 * 2$

           $= 68$

Now, in the triangle ABC,

$\angle A + \angle B + \angle C = 180$

$\angle A + 68 + 68 = 180$

$\angle A = 44$

Therefore, the angle A is 44°.

Although a figure of your question is missing, you might be referring to the one attached.