Question

The external bisectors of $\angle B$ and $\angle C$meet at O . If $\angle A$ is equal to ${50^ \circ }$ ,then the magnitude of $\angle BOC$ is
a.${140^ \circ }$
b.${105^ \circ }$
c.${65^ \circ }$
d.${60^ \circ }$

Answer 1

Hint: Considering the angles $\angle A$,$\angle B$ and $\angle C$to be the exterior angles of the triangle ABC and using the property the sum of the exterior angles of any closed figure is ${360^ \circ }$we get that $\angle B + \angle C = {230^ \circ }$ and then considering the triangle BOC and using the property the sum of the angles of a triangle is ${180^ \circ }$ we get the required angle.

Complete step-by-step answer:
Let's consider a triangle ABC
And let $\angle A$,$\angle B$ and $\angle C$be the exterior angles of the triangle
And the bisectors of $\angle B$ and $\angle C$meet at O
Now we are given $\angle A$ is equal to ${50^ \circ }$ and this mentions the interior angle

Hence the exterior angle is given by subtracting it from ${180^ \circ }$ as the angle formed by straight line is ${180^ \circ }$
$ \Rightarrow \angle A = 180 - 50 = {130^ \circ }$

We know that the sum of the exterior angles of any closed figure is ${360^ \circ }$
$
   \Rightarrow \angle A + \angle B + \angle C = {360^ \circ } \\
   \Rightarrow {130^ \circ } + \angle B + \angle C = {360^ \circ } \\
   \Rightarrow \angle B + \angle C = {360^ \circ } - {130^ \circ } \\
   \Rightarrow \angle B + \angle C = {230^ \circ } \\
$
Now lets consider the triangle BOC
Since $\angle B$ is bisected we get $\angle OBC = \dfrac{{\angle B}}{2}$
Same way , Since $\angle C$ is bisected we get $\angle OCB = \dfrac{{\angle C}}{2}$
We know that the sum of the angles of the triangle is ${180^ \circ }$
$
   \Rightarrow \angle OBC + \angle OCB + \angle BOC = {180^ \circ } \\
   \Rightarrow \dfrac{{\angle B}}{2} + \dfrac{{\angle C}}{2} + \angle BOC = {180^ \circ } \\
   \Rightarrow \dfrac{{\angle B + \angle C}}{2} + \angle BOC = {180^ \circ } \\
$
We know that $\angle B + \angle C = {230^ \circ }$
Using that we get
 $
   \Rightarrow \dfrac{{{{230}^ \circ }}}{2} + \angle BOC = {180^ \circ } \\
   \Rightarrow {115^ \circ } + \angle BOC = {180^ \circ } \\
   \Rightarrow \angle BOC = {180^ \circ } - {115^ \circ } \\
   \Rightarrow \angle BOC = {65^ \circ } \\
$
Hence we get the required angle
The correct option is c
Note: The sum of the length of two sides of a triangle is always greater than the length of the third side.
A triangle with vertices P, Q, and R is denoted as $\vartriangle PQR$ .
The area of a triangle is equal to half of the product of its base and height.