Question

∠5=(3x + 10)°, ∠6=(5x - 50)°. state their angle relationship. solve for x. find the measure of ∠5 and ∠6.

Explanation:

Step1: Assume angles are supplementary

If we assume $\angle5$ and $\angle6$ are supplementary (a common relationship when no other information is given about parallel - lines etc., and we need to solve for $x$), then $\angle5+\angle6 = 180^{\circ}$.
$(3x + 10)+(5x-50)=180$

Step2: Combine like - terms

Combine the $x$ terms and the constant terms on the left - hand side.
$3x+5x+10 - 50=180$
$8x-40 = 180$

Step3: Add 40 to both sides

To isolate the term with $x$, add 40 to both sides of the equation.
$8x-40 + 40=180 + 40$
$8x=220$

Step4: Solve for $x$

Divide both sides by 8.
$x=\frac{220}{8}=\frac{55}{2}=27.5$

Step5: Find the measure of $\angle5$

Substitute $x = 27.5$ into the expression for $\angle5$.
$\angle5=3x + 10=3\times27.5+10=82.5 + 10=92.5^{\circ}$

Step6: Find the measure of $\angle6$

Substitute $x = 27.5$ into the expression for $\angle6$.
$\angle6=5x-50=5\times27.5-50=137.5-50 = 87.5^{\circ}$

Answer:

$x = 27.5$, $\angle5=92.5^{\circ}$, $\angle6=87.5^{\circ}$