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in the figure below, m < pqs = 37°, m < sqr = (6x + 16)°, and m < pqr =…

Question

in the figure below, m < pqs = 37°, m < sqr = (6x + 16)°, and m < pqr = 137°, find x.

Explanation:

Step1: Use angle - addition postulate

We know that $\angle PQR=\angle PQS+\angle SQR$. So, $137 = 37+(6x + 16)$.

Step2: Simplify the right - hand side

First, simplify the right - hand side of the equation: $37+(6x + 16)=6x+37 + 16=6x + 53$. So the equation becomes $137=6x + 53$.

Step3: Isolate the term with $x$

Subtract 53 from both sides of the equation: $137-53=6x+53 - 53$. This gives $84 = 6x$.

Step4: Solve for $x$

Divide both sides of the equation by 6: $\frac{84}{6}=x$. So, $x = 14$.

Answer:

$x = 14$