Chapter 11.2, Problem 50CCR

a.

To determine

To find: Value of $x$ .

Answer to Problem 50CCR

  $x=9^0$

Explanation of Solution

Given information:

  $m\angle ABC=4x^0$

  $m\angle DBE={\left(2x+18\right)}^0$

Angles ABC and DBE are Vertical angles.

Given

Angles ABC and DBE are Vertical angles.

We know Vertical angles are congruent

Therefore,

  $4x^0={\left(2x+18\right)}^0$

On solving

We get

  $\begin{array}{l}4x^0={\left(2x+18\right)}^0\\ \Rightarrow 4x^0=2x^0+{18}^0\\ \Rightarrow 4x^0-2x^0={18}^0\\ \Rightarrow 2x^0={18}^0\\ \Rightarrow \frac{2x^0}{2}=\frac{{18}^0}{2}\\ \Rightarrow x^0=9^0\end{array}$

Therefore

Value of $x$ is

  $x=9^0$

b.

To determine

To find: Value of $m\angle DBE$ .

Answer to Problem 50CCR

  $m\angle DBE={36}^0$

Explanation of Solution

Given information:

  $m\angle ABC=4x^0$

  $m\angle DBE={\left(2x+18\right)}^0$

Angles ABC and DBE are Vertical angles.

Given

Angles ABC and DBE are Vertical angles.

We know Vertical angles are congruent

Therefore,

  $4x^0={\left(2x+18\right)}^0$

On solving

We get

  $\begin{array}{l}4x^0={\left(2x+18\right)}^0\\ \Rightarrow 4x^0=2x^0+{18}^0\\ \Rightarrow 4x^0-2x^0={18}^0\\ \Rightarrow 2x^0={18}^0\\ \Rightarrow \frac{2x^0}{2}=\frac{{18}^0}{2}\\ \Rightarrow x^0=9^0\end{array}$

Therefore

Value of $x$ is

  $x=9^0$

We know

  $m\angle DBE={\left(2x+18\right)}^0$

On substituting values

We get

  $\begin{array}{l}m\angle DBE={\left(2x+18\right)}^0\\ \Rightarrow m\angle DBE={\left(2\left(9^0\right)+18\right)}^0\\ \Rightarrow m\angle DBE={\left(18+18\right)}^0\\ \Rightarrow m\angle DBE={36}^0\end{array}$

Therefore,

  $m\angle DBE={36}^0$