Question

m∠pqs = x² - 5, m∠pqr = 45
m∠rqs = 5x
find m∠pqs.
x=

m∠rqs =

m∠pqs =

Explanation:

Step1: Use angle - addition postulate

Since $\angle PQR=\angle PQS+\angle RQS$, we have the equation $45 = x^{2}-5 + 5x$.

Step2: Rearrange to quadratic form

Rearrange the equation to $x^{2}+5x - 50=0$.

Step3: Factor the quadratic equation

Factor $x^{2}+5x - 50=0$ as $(x + 10)(x - 5)=0$.

Step4: Solve for x

Set each factor equal to zero: $x+10 = 0$ gives $x=-10$ and $x - 5=0$ gives $x = 5$. But since angle measures are non - negative, we discard $x=-10$ and take $x = 5$.

Step5: Find $m\angle RQS$

Substitute $x = 5$ into $m\angle RQS=5x$, so $m\angle RQS=5\times5 = 25^{\circ}$.

Step6: Find $m\angle PQS$

Substitute $x = 5$ into $m\angle PQS=x^{2}-5$, so $m\angle PQS=5^{2}-5=20^{\circ}$.

Answer:

$x = 5$
$m\angle RQS = 25$
$m\angle PQS = 20$