Question

if $overline{ps}congoverline{rs}$, $mangle pqs = 10x$, and $mangle rqs=x + 63^{circ}$, what is $mangle pqr$?
$mangle pqr=square^{circ}$

Explanation:

Step1: Set equal the two angle - measures

Since $\overline{PS}\cong\overline{RS}$ and $\angle{PSQ}=\angle{RSQ} = 90^{\circ}$ and $SQ$ is common, triangles $\triangle{PSQ}$ and $\triangle{RSQ}$ are congruent by the Hypotenuse - Leg (HL) congruence criterion. So, $\angle{PQS}=\angle{RQS}$. Set $10x=x + 63$.
$10x=x + 63$

Step2: Solve for x

Subtract $x$ from both sides of the equation:
$10x−x=x + 63−x$
$9x=63$
Divide both sides by 9:
$x=\frac{63}{9}=7$

Step3: Find the measure of $\angle{PQR}$

Since $\angle{PQR}=\angle{PQS}+\angle{RQS}$ and $\angle{PQS}=\angle{RQS}$, and $\angle{PQS}=10x$ and $\angle{RQS}=x + 63$. Substitute $x = 7$ into either expression for $\angle{PQS}$ or $\angle{RQS}$ and then double it.
$\angle{PQS}=10x=10\times7 = 70^{\circ}$
$\angle{RQS}=x + 63=7+63 = 70^{\circ}$
$\angle{PQR}=2\times70^{\circ}=140^{\circ}$

Answer:

$140$