Question

In isosceles triangle ABC above, $\overline{AM}$ and $\overline{CM}$ are the angle bisectors of angle BAC and angle BCA. What is the measure of angle AMC ?

• A. ${110}^o$
• B. ${115}^o$
• C. ${120}^o$
• D. ${125}^o$
• E. ${130}^o$

Answer 1

Answer: (A)

${110}^o$

Given that

$\angle$$ABC$ $=$ $40$

$△$$ABC$ is an isosceles triangle.

We know that in an isosceles triangle, $base$ angles are equal.

Here, $\angle$$BAC$ and $\angle$$BCA$ are equal.

Let $\angle$$BAC$ $=$ $\angle$$BCA$ $=$ $x$

In $△$$ABC$, $\angle$$BAC$ $+$ $\angle$$BCA$ $+$ $\angle$$ABC$ $=$ $180$

From given and above,

$x$ $+$ $x$ $+$ $40$ $=$ $180$

$2x$ $+$ $40$ $=$ $180$

$2x$ $=$ $140$

$x$ $=$ $70$

Therefore, $\angle$$BAC$ $=$ $\angle$$BCA$ $=$ $70$

As $\overrightarrow{AM}$ and $\overrightarrow{CM}$ are the angle bisectors of the $\angle$$BAC$ and $\angle$$BCA$ respectively.

An $angular$ $bisector$ divides the angle into two equal halves.

Hence, $\angle$$MAC$ $=$ $\angle$$MCA$ $=$ $\frac{x}{2}$

In $△$$AMC$, $\angle$$MAC$ $+$ $\angle$$MCA$ $+$ $\angle$$AMC$ $=$ $180$

From above,

$\frac{x}{2}$ $+$ $\frac{x}{2}$ $+$ $\angle$$AMC$ $=$ $180$

$x$ $+$ $\angle$$AMC$ $=$ $180$

$\angle$$AMC$ $=$ $180$ $-$ $x$

$\angle$$AMC$ $=$ $180$ $-$ $70$

$\angle$$AMC$ $=$ $110$

Therefore, the measure of $\angle$$AMC$ is ${}^{\prime }$${110}^0$${}^{\prime }$.