Question

Construct $△ABC$ in which $BC=7.2cm$, $\angle A={45}^{\circ }$. $M$ is the midpoint of $BC$ and $AM$ is the median of length $8cm$. Then $m\angle B+m\angle C=$

• A. ${125}^{\circ }$
• B. ${135}^{\circ }$
• C. ${145}^{\circ }$
• D. ${155}^{\circ }$

Answer 1

The correct option is B ${135}^{\circ }$

Draw a line segment $BC=7.2cm$. Construct $\angle CBD={45}^o$
At $B$ on $BD$, construct a perpendicular.

Now, draw perpendicular bisector of $BC$, which intersects the perpendicular of $BD$

Taking $O$ as a center and $OB$ as radius, draw a circle

This circle passes through $B$ and $C$. $BD$ is tangent

$M$ as a center, draw an arc of $8cm$ on a circle. Arc cuts the circle at $A$

Join $A,C$ and $A,B$

$ABC$ is required a triangle
In $△ABC,$

$\angle ABC+\angle BAC+\angle ACB={180}^o$

$\angle ABC+\angle ACB+{45}^o={180}^o$

$\therefore m\angle B+m\angle C={135}^o$