Question

∠efg and ∠gfh are a linear pair, m∠efg = 5n + 15, and m∠gfh = 2n + 11. what are m∠efg and m∠gfh?
m∠efg = (square^{circ})
m∠gfh = (square^{circ})
(simplify your answers.)

Explanation:

Step1: Use linear - pair property

Since $\angle EFG$ and $\angle GFH$ are a linear pair, $m\angle EFG + m\angle GFH=180^{\circ}$. So, $(5n + 15)+(2n + 11)=180$.

Step2: Combine like - terms

Combine the $n$ terms and the constant terms: $5n+2n+15 + 11=180$, which simplifies to $7n+26 = 180$.

Step3: Solve for $n$

Subtract 26 from both sides: $7n=180 - 26=154$. Then divide both sides by 7: $n=\frac{154}{7}=22$.

Step4: Find $m\angle EFG$

Substitute $n = 22$ into the expression for $m\angle EFG$: $m\angle EFG=5n + 15=5\times22+15=110 + 15=125^{\circ}$.

Step5: Find $m\angle GFH$

Substitute $n = 22$ into the expression for $m\angle GFH$: $m\angle GFH=2n + 11=2\times22+11=44 + 11=55^{\circ}$.

Answer:

$m\angle EFG = 125^{\circ}$
$m\angle GFH = 55^{\circ}$