Question

1. (10 pts) use the figure below to solve for \\(x\\) and find \\(m\angle rpq\\) and \\(m\angle prq\\)

Explanation:

Identify the geometric relationship

The figure shows a triangle \(\triangle PRQ\) with an extended side forming a line \(PH\). The angle \(\angle QRH\) is an exterior angle of \(\triangle PRQ\) at vertex \(R\), with a measure of \(7x^\circ\). The two remote interior angles are \(\angle RPQ = (x^2 - 6x)^\circ\) and \(\angle PQR = 40^\circ\). By the Exterior Angle Theorem, the measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.

Set up the equation

Using the Exterior Angle Theorem, we write the equation:

$$ m\angle QRH = m\angle RPQ + m\angle PQR $$

Substituting the given expressions:

$$ 7x = (x^2 - 6x) + 40 $$

Solve the quadratic equation

Using the Quadratic Equation Solving concept, we rearrange the equation into standard form \(ax^2 + bx + c = 0\):

$$ x^2 - 13x + 40 = 0 $$

We factor the quadratic expression:

$$ (x - 8)(x - 5) = 0 $$

This gives two potential solutions for \(x\):

$$ x = 8 \quad \text{or} \quad x = 5 $$

Verify the solutions

We must check if both values of \(x\) yield positive angle measures.

• Case 1: If \(x = 5\):

$$ m\angle RPQ = 5^2 - 6(5) = 25 - 30 = -5^\circ $$

An angle measure cannot be negative, so \(x = 5\) is extraneous.

• Case 2: If \(x = 8\):

$$ m\angle RPQ = 8^2 - 6(8) = 64 - 48 = 16^\circ $$

$$ m\angle QRH = 7(8) = 56^\circ $$

Both measures are positive and geometrically valid. Thus, \(x = 8\).

Calculate the required angle measures

Using \(x = 8\), we find the requested angle measures:

• For \(m\angle RPQ\):

$$ m\angle RPQ = 16^\circ $$

• For \(m\angle PRQ\):

Since \(\angle PRQ\) and \(\angle QRH\) form a linear pair:

$$ m\angle PRQ = 180^\circ - m\angle QRH = 180^\circ - 56^\circ = 124^\circ $$

Answer:

• \(x = 8\)
• \(m\angle RPQ = 16^\circ\)
• \(m\angle PRQ = 124^\circ\)