Question

In the given figure, AM = AD, $\angle B={{63}^{\circ }}$ and CD is an angle bisector of $\angle C$, then $\angle MAC=?$

(a) ${{27}^{\circ }}$
(b) ${{37}^{\circ }}$
(c) ${{63}^{\circ }}$
(d) None of these.

Answer 1

Hint: Assume $\angle ACD=\angle BCD=x$ and $\angle AMD=\angle ADM=y$. Also assume the angle MAC, which we have to find, as ‘k’. Use the property of a triangle that ‘sum of the internal angles of a triangle is 180 degrees. Form an equation in $x\ \text{and }y$. Now, in triangle ACM use the property that ‘an exterior angle of a triangle is equal to the sum of opposite interior angles. Form another equation relating $x\text{ and }y$, hence, solve the two equations to get the answer.

Complete step by step answer:
Let us assume that, $\angle ACD=\angle BCD=x$, since it is given that CD is the angle bisector of $\angle C$.
Now, it is given that AM = AD. Therefore, we can conclude that triangle AMD is an isosceles triangle.
Therefore assume, $\angle AMD=\angle ADM=y$.
Now, assume angle MAC is ‘k’. We have to find the value of ‘k’.

Since, $\angle ADM=y$, therefore by linear pair $\angle BDC={{180}^{\circ }}-y$.
Therefore, in triangle BDC,
$\begin{align}
  & x+{{180}^{\circ }}-y+{{63}^{\circ }}={{180}^{\circ }} \\
 & \Rightarrow x-y+{{63}^{\circ }}=0 \\
 & \Rightarrow y-x={{63}^{\circ }}......................(i) \\
\end{align}$
Now, in triangle AMC, using the exterior angle property which says “an exterior angle of a triangle is equal to the sum of opposite interior angles”, we get,
$\begin{align}
  & x+k=y \\
 & \Rightarrow y-x=k........................(ii) \\
\end{align}$
From equations (i) and (ii), we get,
$k={{63}^{\circ }}$

So, the correct answer is “Option C”.

Note: We have used two basic properties of a triangle to find the required angle. This is the easiest approach to solve this question. One may note that we can also solve this problem by the help of trigonometry but that will be lengthier and also we will require the help of a calculator. So, it is better to use the properties of the triangle.