QUESTION IMAGE
Question
- ∠qrs measures 10 more than triple its complement. what is the measure of the supplement of ∠qrs?
a 20°
b 70°
c 110°
d 160°
Step1: Define the complement relationship
Let the measure of $\angle QRS$ be $x$ and its complement be $y$. We know that $x + y=90^{\circ}$, so $y = 90 - x$.
Step2: Set up the equation based on the problem
The problem states that $x=3y + 10$. Substitute $y = 90 - x$ into this equation:
$x=3(90 - x)+10$.
Step3: Solve the equation for $x$
Expand the right - hand side: $x = 270-3x + 10$.
Add $3x$ to both sides: $x+3x=270 + 10$, so $4x=280$.
Divide both sides by 4: $x = 70^{\circ}$.
Step4: Find the supplement of $\angle QRS$
Let the supplement of $\angle QRS$ be $z$. We know that $x+z = 180^{\circ}$. Since $x = 70^{\circ}$, then $z=180 - 70=110^{\circ}$.
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C. $110^{\circ}$