Question

Construct a triangle $ABC$ having given $BC=75$ cm, $\angle B={60}^0$ and $\angle C={45}^0.$ Draw the bisectors of $\angle B$ and $\angle C.$ If these bisectors meet at $I$, find the measure of $\angle BIC$.

• A. ${90}^o$
• B. ${120}^o$
• C. ${127.5}^o$
• D. ${145}^o$

Answer 1

The correct option is C ${127.5}^o$

Draw a line segment $BC=75cm$

Make $\angle B={60}^o$ and $\angle C={45}^o$

Bisect the angles $B$ and $C$ and name the intersection point of bisectors as $I$.

Now, the points $B,C$ and $I$ form a triangle.

In $△BCI$, $\angle B=\frac{60}{2}={30}^o$, $\angle C=\frac{45}{2}={22.5}^o$

We know that sum of three angles of a triangle is ${180}^o$

$\therefore \angle B+\angle C+\angle I={180}^o$

$⟹30+22.5+\angle I={180}^o$

$⟹\angle I=180-52.5={127.5}^o$

Hence, $\angle BIC={127.5}^o$