Question

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

Answer 1

Hint: Considering the angles $\mathrm{\angle }A$,$\mathrm{\angle }B$ and $\mathrm{\angle }C$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 $\mathrm{\angle }B+\mathrm{\angle }C={230}^{\circ }$ 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 $\mathrm{\angle }A$,$\mathrm{\angle }B$ and $\mathrm{\angle }C$be the exterior angles of the triangle
And the bisectors of $\mathrm{\angle }B$ and $\mathrm{\angle }C$meet at O
Now we are given $\mathrm{\angle }A$ 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 ${180}^{\circ }$
$\Rightarrow \mathrm{\angle }A=180-50={130}^{\circ }$

We know that the sum of the exterior angles of any closed figure is ${360}^{\circ }$
$$\begin{gathered} \Rightarrow \mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={360}^{\circ } \\ \Rightarrow {130}^{\circ }+\mathrm{\angle }B+\mathrm{\angle }C={360}^{\circ } \\ \Rightarrow \mathrm{\angle }B+\mathrm{\angle }C={360}^{\circ }-{130}^{\circ } \\ \Rightarrow \mathrm{\angle }B+\mathrm{\angle }C={230}^{\circ } \end{gathered}$$
Now lets consider the triangle BOC
Since $\mathrm{\angle }B$ is bisected we get $\mathrm{\angle }OBC={\displaystyle \frac{\mathrm{\angle }B}{2}}$
Same way , Since $\mathrm{\angle }C$ is bisected we get $\mathrm{\angle }OCB={\displaystyle \frac{\mathrm{\angle }C}{2}}$
We know that the sum of the angles of the triangle is ${180}^{\circ }$
$$\begin{gathered} \Rightarrow \mathrm{\angle }OBC+\mathrm{\angle }OCB+\mathrm{\angle }BOC={180}^{\circ } \\ \Rightarrow {\displaystyle \frac{\mathrm{\angle }B}{2}}+{\displaystyle \frac{\mathrm{\angle }C}{2}}+\mathrm{\angle }BOC={180}^{\circ } \\ \Rightarrow {\displaystyle \frac{\mathrm{\angle }B+\mathrm{\angle }C}{2}}+\mathrm{\angle }BOC={180}^{\circ } \end{gathered}$$
We know that $\mathrm{\angle }B+\mathrm{\angle }C={230}^{\circ }$
Using that we get
 $$\begin{gathered} \Rightarrow {\displaystyle \frac{{230}^{\circ }}{2}}+\mathrm{\angle }BOC={180}^{\circ } \\ \Rightarrow {115}^{\circ }+\mathrm{\angle }BOC={180}^{\circ } \\ \Rightarrow \mathrm{\angle }BOC={180}^{\circ }-{115}^{\circ } \\ \Rightarrow \mathrm{\angle }BOC={65}^{\circ } \end{gathered}$$
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 $△PQR$ .
The area of a triangle is equal to half of the product of its base and height.