Question

1. find m∠deh and m∠feh (10x + 21) 13x

Explanation:

Step1: Set up the equation

Since $\angle DEH+\angle FEH=\angle DEF$ and $\angle DEF = 90^{\circ}$, we have $13x+(10x + 21)=90$.

Step2: Combine like terms

$13x+10x+21=23x + 21$, so $23x+21 = 90$.

Step3: Isolate the variable term

Subtract 21 from both sides: $23x=90 - 21=69$.

Step4: Solve for x

Divide both sides by 23: $x=\frac{69}{23}=3$.

Step5: Find $m\angle DEH$

Substitute $x = 3$ into the expression for $\angle DEH$: $m\angle DEH=13x=13\times3 = 39^{\circ}$.

Step6: Find $m\angle FEH$

Substitute $x = 3$ into the expression for $\angle FEH$: $m\angle FEH=10x + 21=10\times3+21=51^{\circ}$.

Answer:

Let's assume that $\angle DEH = 13x$ and $\angle FEH=10x + 21$, and $\angle DEF = 90^{\circ}$ (since it looks like a right - angle from the figure). Then $\angle DEH+\angle FEH=\angle DEF$.

So, $13x+(10x + 21)=90$.

First, combine like terms:
$13x+10x+21 = 90$
$23x+21 = 90$.

Subtract 21 from both sides:
$23x=90 - 21$
$23x=69$.

Divide both sides by 23:
$x = 3$.

Now find $\angle DEH$:
$\angle DEH=13x=13\times3 = 39^{\circ}$.

Find $\angle FEH$:
$\angle FEH=10x + 21=10\times3+21=30 + 21=51^{\circ}$.

So, $m\angle DEH = 39^{\circ}$ and $m\angle FEH = 51^{\circ}$.