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3. if $mangle efh=(5x + 1)^{circ}$, $mangle hfg = 62^{circ}$, and $mang…

Question

  1. if $mangle efh=(5x + 1)^{circ}$, $mangle hfg = 62^{circ}$, and $mangle efg=(18x + 11)^{circ}$, find each measure.

x =
$mangle efh=$
$mangle efg=$

Explanation:

Step1: Use angle - addition postulate

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

Step2: Simplify the right - hand side of the equation

$(5x + 1)+62=5x+63$. So the equation becomes $18x + 11=5x+63$.

Step3: Solve for $x$

Subtract $5x$ from both sides: $18x-5x + 11=5x-5x+63$, which simplifies to $13x+11 = 63$. Then subtract 11 from both sides: $13x+11 - 11=63 - 11$, giving $13x=52$. Divide both sides by 13: $x=\frac{52}{13}=4$.

Step4: Find $m\angle EFH$

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

Step5: Find $m\angle EFG$

Substitute $x = 4$ into the expression for $m\angle EFG$. $m\angle EFG=(18x + 11)=(18\times4+11)=72 + 11=83^{\circ}$.

Answer:

$x = 4$
$m\angle EFH=21^{\circ}$
$m\angle EFG=83^{\circ}$