Question

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

$\mathit{m}\mathbf{\angle }\mathbf{3}\mathbf{=}\mathbf{6}\mathit{x}$,$\mathit{m}\mathbf{\angle }\mathit{K}\mathit{A}\mathit{T}\mathbf{=}\mathbf{90}\mathbf{-}\mathit{x}$

Short answer

The value of$\mathit{x}$is ${\mathbf{18}}^{\mathbf{\text{o}}}$.

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$SAT$ is a straight line, therefore,

$m\angle 3+m\angle KAT={180}^{°}$

- Calculate x.

In order to calculate the value of $x$, substitute $6x$for $m\angle 3$and $90-x$for $m\angle KAT$into the equation $m\angle 3+m\angle KAT={180}^{°}$and simplify.

$\begin{array}{c}6x+{90}^{°}-x={180}^{°}\\ 5x+{90}^{°}={180}^{°}\end{array}$

Subtract ${90}^{°}$from both sides.

$\begin{array}{c}5x+{90}^{°}-{90}^{°}={180}^{°}-{90}^{°}\\ 5x={90}^{°}\\ x=\frac{{90}^{°}}{5}\\ x={18}^{°}\end{array}$

Therefore, $x={18}^{°}$.