Question

In the given figure, BO, CO are the angle bisectors of external angles of $$\triangle ABC$$. Then $$\angle BOC$$______.

A. $$90^{\circ }-\dfrac{1}{2} \angle A$$
B. $$90^{\circ }+\dfrac{1}{2} \angle A$$
C. $$180^{\circ }-\dfrac{1}{2} \angle A$$
D. $$180^{\circ }+\dfrac{1}{2} \angle A$$

Answer 1

Hint: In this question it is given that in the given figure, BO, CO are the angle bisectors of external angles of $$\triangle ABC$$. Then we have to find $$\angle BOC$$. So to find the solution we need to know the following two properties
First property: the summation of all the angles of a triangle is $$180^{\circ }$$.
Second property: If two angles are a linear pair (angle on a straight line) then the angles are supplementary, i,e sum of the angles is $$180^{\circ }$$.
So by using the above information we have to find the solution.

Complete step-by-step answer:
The external angles of $$\triangle ABC$$ are $$\angle PBC$$ and $$\angle QCB$$.
Here It is given that BO, CO are the angle bisectors of external angles of $$\triangle ABC$$, i.e, BO and CO is the bisector of $$\angle PBC$$ and $$\angle QCB$$.
Since the created angles by BO is $$\angle 1$$ and $$\angle 2$$
$$\therefore \angle 1=\angle 2$$
Let, $$ \angle 1=\angle 2=x$$
Similarly CO is the bisector of $$\angle QCB$$
Therefore we can write from the diagram,
$$ \angle 3=\angle 4$$
Let, $$ \angle 3=\angle 4=y$$
Since $$\angle ABC$$ and $$\angle PBC$$ are the linear pairs,
Therefore by the second property we can write,
$$\angle ABC+\angle PBC=180^{\circ }$$
$$\Rightarrow \angle ABC+\angle 1+\angle 2=180^{\circ }$$
$$\Rightarrow \angle ABC+x+x=180^{\circ }$$
$$\Rightarrow \angle ABC+2x=180^{\circ }$$
$$\Rightarrow \angle ABC=180^{\circ }-2x$$..........(1)
Again since $$\angle ACB$$ and $$\angle QCB$$ are the linear pairs, so similarly by the above process we can write,
$$\angle ACB=180^{\circ }-2y$$.........(2)
Now by using the triangular formulas(first property) we can write for $$\triangle ABC$$
$$\angle A+\angle ABC+\angle ACB=180^{\circ }$$ [since,$$\angle BAC =\angle A$$]
$$\Rightarrow \angle A+\left( 180^{\circ }-2x\right) +\left( 180^{\circ }-2y\right) =180^{\circ }$$[by (1) and (2)]
$$\Rightarrow \angle A+180^{\circ }+180^{\circ }-2x-2y=180^{\circ }$$
$$\Rightarrow \angle A+360^{\circ }=180^{\circ }+2x+2y$$
$$\Rightarrow 180^{\circ }+2x+2y=\angle A+360^{\circ }$$
$$\Rightarrow 2x+2y=\angle A+360^{\circ }-180^{\circ }$$
$$\Rightarrow 2(x+y)=\angle A+180^{\circ }$$
$$\Rightarrow x+y=\dfrac{\angle A+180^{\circ }}{2}$$
$$\Rightarrow x+y=\dfrac{180^{\circ }}{2} +\dfrac{1}{2} \angle A$$
$$\Rightarrow x+y=90^{\circ }+\dfrac{1}{2} \angle A$$.........(3)

Now again using the triangular formulas(first property) we can write for $$\triangle BOC$$
$$\angle BOC+\angle 4+\angle 1=180^{\circ }$$
$$\Rightarrow \angle BOC+y+x=180^{\circ }$$
$$\Rightarrow \angle BOC=180^{\circ }-x-y$$
$$\Rightarrow \angle BOC=180^{\circ }-\left( x+y\right) $$
Now by putting the value of (x+y) from equation (3), we can write the above equation as,
$$ \angle BOC=180^{\circ }-\left( 90^{\circ }+\dfrac{1}{2} \angle A\right) $$
$$\Rightarrow \angle BOC=180^{\circ }-90^{\circ }-\dfrac{1}{2} \angle A$$
$$\Rightarrow \angle BOC=90^{\circ }-\dfrac{1}{2} \angle A$$

So, the correct answer is “Option A”.

Note: While solving this type of question you need to know that, if you have given a line which is an angle bisector, then it means that the line bisects that angle, i.e, the line divides the angle in equal two measures.