Question

1. m∠prs = ________°

m∠p = (6x + 9)°
m∠q = (15x + 9)°
m∠prs = (9x²)°

Explanation:

Step1: Use the exterior - angle theorem

The exterior - angle theorem of a triangle states that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non - adjacent interior angles.
For \(\triangle PQR\), \(\angle PRS\) is an exterior angle. So, \(m\angle PRS=m\angle P + m\angle Q\).
Substitute the given angle expressions: \(9x^{2}=(6x + 9)+(15x + 9)\).

Step2: Simplify the right - hand side of the equation

First, combine like terms on the right - hand side: \((6x + 9)+(15x + 9)=6x+15x + 9 + 9=21x + 18\).
So, the equation becomes \(9x^{2}-21x - 18 = 0\).
Divide the entire equation by \(3\) to simplify: \(3x^{2}-7x - 6 = 0\).

Step3: Solve the quadratic equation \(3x^{2}-7x - 6 = 0\)

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 3\), \(b=-7\), and \(c=-6\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-7)^{2}-4\times3\times(-6)=49 + 72=121\).
Then, \(x=\frac{7\pm\sqrt{121}}{6}=\frac{7\pm11}{6}\).
We get two solutions:
\(x_1=\frac{7 + 11}{6}=\frac{18}{6}=3\) and \(x_2=\frac{7-11}{6}=\frac{-4}{6}=-\frac{2}{3}\).
Since \(x\) represents an angle - related value, we discard \(x =-\frac{2}{3}\) (because if \(x =-\frac{2}{3}\), \(m\angle P=6x + 9=6\times(-\frac{2}{3})+9=-4 + 9 = 5\), \(m\angle Q=15x + 9=15\times(-\frac{2}{3})+9=-10 + 9=-1\), and angle measures cannot be negative in this context).

Step4: Find the measure of \(\angle PRS\)

Substitute \(x = 3\) into \(m\angle PRS=9x^{2}\).
\(m\angle PRS=9\times3^{2}=9\times9 = 81\).

Answer:

\(81\)