Question

for questions #7 - 9, refer to the diagram at the right, where pq⊥qr, m∠pqt = 2x - 5, m∠tqs = 4x + 5, and m∠sqr = 3x.

1. find m∠pqr.
2. find m∠tqr.
3. find the complement of ∠pqs in degrees.

Explanation:

Step1: Use angle - addition property

Since $\overline{PQ}\perp\overline{QR}$, $\angle PQR = 90^{\circ}$. Also, $\angle PQT+\angle TQR=\angle PQR$. And $\angle TQS+\angle SQR=\angle TQR$.

Step2: Find the value of $x$

We know that $\angle PQT = 2x - 5$, $\angle TQS=4x + 5$, and $\angle SQR = 3x$. Since $\angle PQR = 90^{\circ}$, and $\angle PQT+\angle TQS+\angle SQR=\angle PQR$, we substitute the expressions: $(2x - 5)+(4x + 5)+3x=90$.
Simplify the left - hand side: $2x-5 + 4x+5+3x=9x$. So, $9x = 90$, and $x = 10$.

Step3: Solve for $\angle PQR$

Since $\overline{PQ}\perp\overline{QR}$, $m\angle PQR=90^{\circ}$.

Step4: Solve for $\angle TQR$

$m\angle TQR=m\angle TQS + m\angle SQR$. Substitute $x = 10$ into the expressions for $\angle TQS$ and $\angle SQR$. $m\angle TQS=4x + 5=4\times10+5 = 45^{\circ}$, $m\angle SQR=3x=3\times10 = 30^{\circ}$. So, $m\angle TQR=45^{\circ}+30^{\circ}=75^{\circ}$.

Step5: Solve for $\angle PQS$

$m\angle PQS=m\angle PQT + m\angle TQS$. $m\angle PQT=2x - 5=2\times10-5 = 15^{\circ}$, $m\angle TQS=4x + 5=45^{\circ}$. So, $m\angle PQS=15^{\circ}+45^{\circ}=60^{\circ}$. The complement of $\angle PQS$ is $90^{\circ}-m\angle PQS$. So the complement is $90 - 60=30^{\circ}$.

Answer:

1. $m\angle PQR = 90^{\circ}$
2. $m\angle TQR = 75^{\circ}$
3. The complement of $\angle PQS$ is $30^{\circ}$