Question

suppose that $\triangle qrs$ is isosceles with base $\overline{rs}$. suppose also that $m\angle r=(5x - 1)\degree$ and $m\angle s=(3x + 21)\degree$. find the degree measure of each angle in the triangle.

Explanation:

Step1: Identify equal angles

In isosceles triangle \( \triangle QRS \) with base \( RS \), the equal sides are \( QR \) and \( QS \), so the base angles are \( \angle R \) and \( \angle S \)? Wait, no—wait, base is \( RS \), so the legs are \( QR \) and \( QS \), so the base angles are \( \angle R \) and \( \angle S \)? Wait, no, in an isosceles triangle, the angles opposite the equal sides are equal. Wait, the base is \( RS \), so the two equal sides are \( QR \) and \( QS \) (since the marks on \( QR \) and \( QS \) indicate they are equal). Therefore, the angles opposite them: \( \angle S \) is opposite \( QR \), and \( \angle R \) is opposite \( QS \). Wait, no—wait, side \( QR \) is opposite angle \( S \), and side \( QS \) is opposite angle \( R \). Since \( QR = QS \), then \( \angle R = \angle S \). Wait, the problem says \( m\angle R = (5x - 1)^\circ \) and \( m\angle S = (3x + 21)^\circ \). Wait, but if \( QR = QS \), then \( \angle R = \angle S \). So set \( 5x - 1 = 3x + 21 \).

Step2: Solve for \( x \)

\( 5x - 1 = 3x + 21 \)
Subtract \( 3x \) from both sides: \( 2x - 1 = 21 \)
Add 1 to both sides: \( 2x = 22 \)
Divide by 2: \( x = 11 \)

Step3: Find \( m\angle R \) and \( m\angle S \)

Substitute \( x = 11 \) into \( m\angle R \): \( 5(11) - 1 = 55 - 1 = 54^\circ \)
Substitute \( x = 11 \) into \( m\angle S \): \( 3(11) + 21 = 33 + 21 = 54^\circ \)

Step4: Find \( m\angle Q \)

The sum of angles in a triangle is \( 180^\circ \). So \( m\angle Q + m\angle R + m\angle S = 180^\circ \)
\( m\angle Q + 54 + 54 = 180 \)
\( m\angle Q + 108 = 180 \)
Subtract 108: \( m\angle Q = 72^\circ \)

Answer:

\( m\angle Q = 72^\circ \), \( m\angle R = 54^\circ \), \( m\angle S = 54^\circ \)