Question

type 3 - angle problems (complement / supplement / linear pairs)
problem 5:
∠lmn measures 12 more than triple its complement. find the measure of the supplement of ∠lmn.

Explanation:

Step1: Define the complement relationship

Let the measure of $\angle LMN$ be $x$ and its complement be $y$. We know that $x + y=90^{\circ}$, so $y = 90 - x$. Also, given that $x=3y + 12$.

Step2: Substitute the expression for $y$ into the second - equation

Substitute $y = 90 - x$ into $x=3y + 12$. We get $x=3(90 - x)+12$.
Expand the right - hand side: $x = 270-3x + 12$.

Step3: Solve for $x$

Add $3x$ to both sides of the equation: $x+3x=270 + 12$.
$4x=282$.
Divide both sides by 4: $x=\frac{282}{4}=70.5^{\circ}$.

Step4: Find the supplement of $x$

Let the supplement of $\angle LMN$ be $z$. We know that $x + z=180^{\circ}$.
So $z=180 - x$. Substitute $x = 70.5^{\circ}$ into the equation, then $z=180 - 70.5=109.5^{\circ}$.

Answer:

$109.5^{\circ}$