Question

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

Answer 1

Answer:

The value of $\angle A$ is 44.

Step-by-step explanation:

Given $\angle BOC =112$.

Since $BO$ bisects $\angle B$, $\angle ABO= \angle OBC$

Also, $CO$ bisects $\angle C$. So $\angle ACO=\angle OCB$

Since in  $\triangle OBC$ , sum of angles is 180, We have

$\angle BOC+\angle OBC+\angle OCB=180\\112+\angle OBC+\angle OCB=180\\\angle OBC+\angle OCB=180-112= 68$

Also in quadrilateral  $ABOC$, the angle sum is 360. That is

$\angle A+\angle ABO+\angle ACO+\angle BOC=360\\\angle A+\angle OBC+\angle OCB+360-112=360\\\angle A+ 68=112\\\\\angle A=112-68= 44$

Here we have used the fact that angle around a point is 360. Also the equality of angles given in first steps is substituted.

Answer 2

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

To find: The angle A.

Solution:

The sum of all angles of a triangle must be 180°. In triangle BOC, let ∠B and ∠C be equal to x° each. This can be written in the form of an equation as follows.

$112 + x + x = 180$

$2x = 68$

$x = 34$

As mentioned in the question, BO and CO are bisectors of angles B and C, respectively. So, ∠ABC is twice ∠OBC and similarly, ∠ACB is twice ∠OCB.

$\angle ABC = 2 * 34$

           $= 68$

$\angle ACB = 2 * 34$

           $= 68$

Now, the sum of all the angles in the triangle ABC can be written as follows.

$68 + 68 + \angle A = 180$

$\angle A = 44$

Therefore, the angle A is 44°.