Question

In each of the following figure, find the value of $ x $ .

Answer 1

Hint: We have isosceles triangles in both figures. So, we will use the properties of the isosceles triangle. We will use the fact that the sum of all angles of a triangle is $ 180{}^\circ $. We know that the sum of the interior angles is equal to the exterior angle. We will use these relations to find the missing angles. This will help us calculate the value of $ x $ in each figure.

Complete step by step answer:
Let us look at figure 1. We are given that $ \angle \text{BAC}=40{}^\circ $ and $ \text{AB}=\text{AC} $ . Therefore, $ \Delta \text{ABC} $ is an isosceles triangle. In an isosceles triangle, the angles opposite to the equal sides are equal. Hence, $ \angle \text{ABC}=\angle \text{ACB} $ . Now, we know that the sum of all angles of a triangle is $ 180{}^\circ $ . So, in $ \Delta \text{ABC} $ , we have the following,
 $ \angle \text{ABC}+\angle \text{ACB }+\angle \text{BAC}=180{}^\circ $
Substituting the value of the known angle and using the relation $ \angle \text{ABC}=\angle \text{ACB} $ , we get
 $ 2\angle \text{ABC}+40{}^\circ =180{}^\circ $
Solving the above equation, we get the following,
 $ \begin{align}
  & 2\angle \text{ABC}=140{}^\circ \\
 & \therefore \angle \text{ABC}=70{}^\circ \\
\end{align} $
We have to find the value of $ \angle \text{ACD} $ . This angle is an exterior angle. Therefore, its value is equal to the sum of interior angles. So, we have
 $ \angle \text{ACD}=\angle \text{BAC}+\angle \text{ABC} $
Substituting the values, we get
 $ \begin{align}
  & x=40{}^\circ +70{}^\circ \\
 & \therefore x=110{}^\circ \\
\end{align} $
Next, we will look at figure 2. We are given that $ \angle \text{ADC}=30{}^\circ $ and $ \text{AC}=\text{CD} $ . Therefore, $ \Delta \text{ACD} $ is an isosceles triangle. In an isosceles triangle, the angles opposite to the equal sides are equal. Hence, $ \angle \text{ACD}=\angle \text{CAD}=30{}^\circ $ .
Now, from the figure, we can see that $ \angle \text{BAD}=\angle \text{BAC}+\angle \text{CAD} $ . We have $ \angle \text{BAC}=65{}^\circ $ and $ \angle \text{CAD}=30{}^\circ $ . Therefore, we get the following,
 $ \begin{align}
  & \angle \text{BAD}=65{}^\circ +30{}^\circ \\
 & \therefore \angle \text{BAD}=95{}^\circ \\
\end{align} $
We will consider $ \Delta \text{ABD} $ . We know that the sum of all angles of a triangle is $ 180{}^\circ $ . So, we have
 $ \angle \text{ABD}+\angle \text{BDA }+\angle \text{BAD}=180{}^\circ $
Substituting the values, we get
 $ \begin{align}
  & x+30{}^\circ +95{}^\circ =180{}^\circ \\
 & \Rightarrow x=180{}^\circ -125{}^\circ \\
 & \therefore x=55{}^\circ \\
\end{align} $

Note:
It is essential that we are familiar with the properties of different triangles. We should also understand the relations between the angles in a given geometric figure. Labeling a figure is very important in order to avoid confusion, especially for the names of angles. In figure 1, we can also use the concept of supplementary angles to find the value of $ x $.