Question

in $\triangle jkl$, $\overline{jl}$ is extended through point $l$ to point $m$, $m\angle jkl = (3x + 8)\degree$, $m\angle ljk = (2x + 20)\degree$, and $m\angle klm = (7x - 10)\degree$. find $m\angle ljk$.

Explanation:

Step1: Use the exterior angle theorem

The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. In \(\triangle JKL\), \(\angle KLM\) is an exterior angle. So, \(m\angle KLM=m\angle JKL + m\angle LJK\).
Substitute the given angle measures: \((7x - 10)=(3x + 8)+(2x + 20)\).

Step2: Simplify the right - hand side of the equation

First, simplify the right - hand side: \((3x + 8)+(2x + 20)=3x+2x + 8 + 20=5x+28\).
The equation becomes \(7x-10 = 5x+28\).

Step3: Solve for \(x\)

Subtract \(5x\) from both sides: \(7x-5x-10=5x - 5x+28\), which gives \(2x-10 = 28\).
Add \(10\) to both sides: \(2x-10 + 10=28+10\), so \(2x=38\).
Divide both sides by \(2\): \(x=\frac{38}{2}=19\).

Step4: Find \(m\angle LJK\)

Since \(m\angle LJK=(2x + 20)^{\circ}\), substitute \(x = 19\) into the expression.
\(m\angle LJK=(2\times19+20)^{\circ}=(38 + 20)^{\circ}=58^{\circ}\).

Answer:

\(58^{\circ}\)