Chapter 4.4, Problem 7WE

To determine

Value of x in the given figure.

Answer to Problem 7WE

The value of $x={41}^o$

Explanation of Solution

Given:

The given figure is,

  

Here, in $\Delta ABD$ and $\Delta BCD$

  $\overline{AD}\cong \overline{CD},\overline{AB}\cong \overline{CB}$ and $\overline{BD}$ is common to both triangles.

So, by SSS postulate

  $\Delta ABD\cong \Delta BCD$

So, $\angle A\cong \angle C$

i.e.

  $m\angle A=m\angle C={98}^o$

Now, consider $\Delta BCD$

  $\overline{CD}\cong \overline{CB}$

So, $\Delta BCD$ is an isosceles triangle.

So, $m\angle CBD=m\angle BDC$

Now,

  $\begin{array}{l}m\angle CBD+m\angle BCD+m\angle BDC=180\\ \Rightarrow x+x+98=180\\ \Rightarrow 2x=180-98\\ \Rightarrow x={41}^o\end{array}$