Question

$\overleftrightarrow{\mathit{A}\mathit{L}}$bisects $\mathit{\angle }\mathit{K}\mathit{A}\mathit{T}$. Find the value of $\mathit{x}$.

$\mathit{m}\mathbf{\angle }\mathbf{1}\mathbf{=}\mathit{x}$,$\mathit{m}\mathbf{\angle }\mathbf{3}\mathbf{=}\mathbf{4}\mathit{x}$

Short answer

The value of$x$ is ${\mathbf{30}}^{\mathbf{\text{o}}}$.

Step-by-step solution

- Line bisector and straight angle

If a line bisects angles, it divides it into two equal parts.

Measure of straight angle is ${180}^{\text{o}}$.

- Observe the figure

The given figure is

From the figure, since $\overleftrightarrow{AL}$bisects $\angle KAT$, therefore, $m\angle 1=m\angle 2=x$.

Also, $\angle SAT$is a straight angel therefore.

$m\angle 1+m\angle 2+m\angle 3={180}^{\text{o}}$

- Calculate x.

In order to calculate the value of $x$, substitute $x$for $m\angle 1$, $x$for $m\angle 2$, $4x$for $m\angle 3$into the equation $m\angle 1+m\angle 2+m\angle 3={180}^{\text{o}}$and simplify.

$\begin{array}{c}x+x+4x={180}^{\text{o}}\\ 6x={180}^{\text{o}}\\ \frac{6x}{6}=\frac{{180}^{\text{o}}}{6}\\ x={30}^{\text{o}}\end{array}$

Therefore, $x={30}^{\text{o}}$.