Question

In the given figure $AD=BD=AC$; $\angle CAE={{75}^{\circ }}$ and $\angle ACD={{x}^{\circ }}$. Find the value of x.

A. ${{45}^{\circ }}$
B. ${{50}^{\circ }}$
C. ${{60}^{\circ }}$
D. $37{{\dfrac{1}{2}}^{\circ }}$

Answer 1

Hint: We assume the value of angles $\angle DAB=\angle DBA$ as y. we use different triangle’s theorems on equal sides and exterior angles. We use them to find the relation between the angles in $\Delta ABD$ and $\Delta ADC$. We get two equations of two unknowns. We solve them to find the value of x and the solution to the problem.

Complete step-by-step solution
We know that in a triangle if two sides are equal in length then the opposite angles of the corresponding sides are also equal.
: In the given figure $AD=BD=AC$.
For the given $\Delta ABD$, $AD=BD$. So, their opposite angles are also equal which means $\angle DAB=\angle DBA$. Let $\angle DAB=\angle DBA=y,y>0$.
For the given $\Delta ADC$, $AD=AC$. So, their opposite angles are also equal which means $\angle ACD=\angle ADC=x$.
We also have the theorem that the exterior angle of a triangle is equal to the sum of the other two interior angles.
For the given $\Delta ABC$, $\angle CAE={{75}^{\circ }}$ is an exterior angle. So, the other two angles are $\angle DBA$ and $\angle ACD$. So, $\angle ACD+\angle DBA=\angle CAE={{75}^{\circ }}$.
Replacing the value of the angles we get $x+y=75.....(i)$.
For the given $\Delta ABD$, $\angle DAB=\angle DBA=y$. $\angle ADC=x$ is an exterior angle. So, the other two angles are $\angle DAB=\angle DBA=y$. So, $\angle DAB+\angle DBA=\angle ADC$.
Replacing the value of the angles we get $y+y={{x}^{\circ }}\Rightarrow 2y=x......(ii)$.
We got two equations of two unknowns. We solve them to find the value of x.
$2y=x\Rightarrow y=\dfrac{x}{2}$. Putting the value in $x+y=75$, we get
$\begin{align}
  & x+y=75 \\
 & \Rightarrow x+\dfrac{x}{2}=75 \\
 & \Rightarrow \dfrac{3x}{2}=75 \\
 & \Rightarrow x=\dfrac{75\times 2}{3}=50 \\
\end{align}$.
Therefore, the value of x is 50.

Note: We have to take care of finding the exterior angles. The exterior angles can be found when any side of a triangle is being extended and that’s why at a particular angle, we can find two exterior angles as the angle has two hands. In the equal hand and equal angles theorem, we need to remember we have to always take the opposite angles of the equal sides.