Question

$AM\perp BC$ and $AN$ is the bisector of $\angle A$. If $\angle ABC={70}^{\circ }$ and $\angle ACB={20}^{\circ }$ find the value of $\angle MAN$

Answer 1

We know that the sum of all the angles in triangle $ABC$ is ${180}^{\circ }$.

$\angle A+\angle B+\angle C={180}^{\circ }$

By substituting the values

$\angle A+{70}^{\circ }+{20}^{\circ }={180}^{\circ }$

On further calculation

$\angle A={180}^{\circ }-{70}^{\circ }-{20}^{\circ }$

By subtraction

$\angle A={180}^{\circ }-{90}^{\circ }$

$\angle A={90}^{\circ }$

We know that the sum of all the angles in triangle $ABM$ is ${180}^{\circ }$.

$\angle BAM+\angle ABM+\angle AMB={180}^{\circ }$

By substituting the values

$\angle BAM+{70}^{\circ }+{90}^{\circ }={180}^{\circ }$

On further calculation

$\angle BAM={180}^{\circ }-{70}^{\circ }-{90}^{\circ }$

By subtraction

$\angle BAM={180}^{\circ }-{160}^{\circ }$

$\angle BAM={20}^{\circ }$

It is given that $AN$ is the bisector of $\angle A$

So it can be written as

$\angle BAN=\frac{1}{2}\angle A$

By substituting the values

$\angle BAN=\frac{1}{2}\left({90}^{\circ }\right)$

By division

$\angle BAN={45}^{\circ }$

From the figure we know that

$\angle MAN+\angle BAM=\angle BAN$

By substituting the values we get

$\angle MAN+{20}^{\circ }={45}^{\circ }$

On further calculation

$\angle MAN={45}^{\circ }-{20}^{\circ }$

By subtraction

$\angle MAN={25}^{\circ }$

Therefore, $\angle MAN={25}^{\circ }$.