Question

1. if (mangle efh=(5x + 1)^{circ}), (mangle hfg = 62^{circ}), and (mangle efg=(18x + 11)^{circ}), find each measure.
2. if (mangle wxz=(5x + 3)^{circ}), (mangle zxy=(8x - 4)^{circ}), and (angle wxy) is a right - angle, find each measure.

Explanation:

Step1: Use angle - addition postulate for first problem

Since $\angle EFG=\angle EFH+\angle HFG$, we have the equation $(18x + 11)=(5x + 1)+62$.

Step2: Simplify the equation

$18x+11 = 5x+63$. Subtract $5x$ from both sides: $18x - 5x+11=5x - 5x+63$, which gives $13x+11 = 63$. Then subtract 11 from both sides: $13x+11 - 11=63 - 11$, so $13x=52$.

Step3: Solve for $x$

Divide both sides of $13x = 52$ by 13: $x=\frac{52}{13}=4$.

Step4: Find $\angle EFH$

Substitute $x = 4$ into the formula for $\angle EFH$: $m\angle EFH=(5x + 1)=(5\times4+1)=21^{\circ}$.

Step5: Find $\angle EFG$

Substitute $x = 4$ into the formula for $\angle EFG$: $m\angle EFG=(18x + 11)=(18\times4+11)=83^{\circ}$.

Step6: Use angle - addition postulate for second problem

Since $\angle WXY$ is a right - angle ($m\angle WXY = 90^{\circ}$) and $\angle WXY=\angle WXZ+\angle ZXY$, we have the equation $(5x + 3)+(8x - 4)=90$.

Step7: Simplify the equation

Combine like terms: $5x+8x+3 - 4=90$, which gives $13x - 1=90$. Add 1 to both sides: $13x-1 + 1=90 + 1$, so $13x=91$.

Step8: Solve for $x$

Divide both sides of $13x = 91$ by 13: $x=\frac{91}{13}=7$.

Step9: Find $\angle WXZ$

Substitute $x = 7$ into the formula for $\angle WXZ$: $m\angle WXZ=(5x + 3)=(5\times7+3)=38^{\circ}$.

Step10: Find $\angle ZXY$

Substitute $x = 7$ into the formula for $\angle ZXY$: $m\angle ZXY=(8x - 4)=(8\times7 - 4)=52^{\circ}$.

Answer:

For problem 3:
$x = 4$
$m\angle EFH=21^{\circ}$
$m\angle EFG=83^{\circ}$
For problem 5:
$x = 7$
$m\angle WXZ=38^{\circ}$
$m\angle ZXY=52^{\circ}$