Question

find the measure of the indicated angle m∠r = type your answer (3x + 1)° (2x + 2)°

Explanation:

Step1: Identify Triangle Type

The triangle \( \triangle RST \) has two equal sides (marked with ticks), so it's isosceles with \( \angle R = \angle T \)? Wait, no—wait, the sides \( RS \) and \( RT \) are equal? Wait, the marks: \( RS \) and \( RT \)? Wait, no, the sides \( SR \) and \( TR \)? Wait, the angle at \( R \) is \( (3x + 1)^\circ \), angle at \( T \) is \( (2x + 2)^\circ \), and since two sides are equal ( \( RS = RT \)? Wait, no, the sides \( SR \) and \( ST \)? Wait, the ticks: one on \( RS \) and one on \( RT \)? Wait, no, the triangle has two sides marked equal, so the base angles are equal. Wait, actually, in an isosceles triangle, the angles opposite equal sides are equal. So if \( RS = RT \), then angles opposite them ( \( \angle T \) and \( \angle S \)) would be equal. But here, angle at \( R \) is \( (3x + 1) \), angle at \( T \) is \( (2x + 2) \). Wait, maybe I misread. Wait, the triangle has \( S \), \( R \), \( T \) as vertices. The sides \( SR \) and \( TR \) are equal? Wait, no, the marks: one on \( SR \) and one on \( ST \)? Wait, the diagram shows two sides with ticks, so \( RS = RT \), so triangle \( RST \) is isosceles with \( RS = RT \), so angles at \( S \) and \( T \) are equal? Wait, no, angle at \( R \) is between \( RS \) and \( RT \), so if \( RS = RT \), then angles at \( S \) and \( T \) are equal. Wait, but the problem gives angle at \( R \) as \( (3x + 1) \) and angle at \( T \) as \( (2x + 2) \). Wait, maybe it's a typo, or maybe \( RS = ST \), so angles at \( R \) and \( T \) are equal? Wait, that must be it. So if \( RS = ST \), then angles opposite them ( \( \angle T \) and \( \angle R \)) are equal? No, angle opposite \( RS \) is \( \angle T \), angle opposite \( ST \) is \( \angle R \). So if \( RS = ST \), then \( \angle T = \angle R \). Wait, that would mean \( 3x + 1 = 2x + 2 \). Let's check that.

Step2: Set Angles Equal (Isosceles Triangle)

Since two sides are equal ( \( RS = ST \) ), the angles opposite them ( \( \angle T \) and \( \angle R \)) are equal. So:
\( 3x + 1 = 2x + 2 \)
Subtract \( 2x \) from both sides:
\( x + 1 = 2 \)
Subtract 1:
\( x = 1 \)

Step3: Calculate \( m\angle R \)

Now, substitute \( x = 1 \) into \( 3x + 1 \):
\( 3(1) + 1 = 3 + 1 = 4 \)? Wait, that can't be right. Wait, maybe I mixed up the angles. Wait, maybe the equal sides are \( SR \) and \( TR \), so angles at \( S \) and \( T \) are equal, and angle at \( R \) is different. Wait, let's re-examine. The triangle has vertices \( S \), \( R \), \( T \). The sides: \( SR \) and \( TR \) are equal (ticks), so \( \angle S = \angle T \). Then angle at \( R \) is \( (3x + 1) \), angle at \( S \) and \( T \) are equal, so \( \angle S = \angle T = (2x + 2) \). Then sum of angles in a triangle is \( 180^\circ \):
\( (3x + 1) + 2(2x + 2) = 180 \)
Simplify:
\( 3x + 1 + 4x + 4 = 180 \)
\( 7x + 5 = 180 \)
\( 7x = 175 \)
\( x = 25 \)
Then \( m\angle R = 3x + 1 = 3(25) + 1 = 75 + 1 = 76^\circ \)
Wait, that makes more sense. So my initial mistake was identifying which angles are equal. Since two sides ( \( SR \) and \( TR \)) are equal, the base angles ( \( \angle S \) and \( \angle T \)) are equal. So angle at \( R \) is \( (3x + 1) \), angles at \( S \) and \( T \) are \( (2x + 2) \) each. Then sum of angles: \( (3x + 1) + 2(2x + 2) = 180 \).

Let's redo Step2 correctly:

Step2: Correct Angle Sum

In triangle \( RST \), \( SR = TR \) (equal sides), so \( \angle S = \angle T = (2x + 2)^\circ \), and \( \angle R = (3x + 1)^\circ \). Sum of angles in a triangle:
\( \angle R + \angle S + \angle T = 180 \)
Substitute:
\( (3x + 1) + (2x + 2) + (2x + 2) = 180 \)
Simplify:
\( 3x + 1 + 2x + 2 + 2x + 2 = 180 \)
\( 7x + 5 = 180 \)
Subtract 5:
\( 7x = 175 \)
Divide by 7:
\( x = 25 \)

Step3: Calculate \( m\angle R \)

Substitute \( x = 25 \) into \( 3x + 1 \):
\( 3(25) + 1 = 75 + 1 = 76 \)

Answer:

\( 76^\circ \)