Question

problem 9: find the measures of the angles of a triangle whose angles have measures of x, 3x, and 3x + 5. also, what kind of triangle is it? (first taught in lesson 30)

Explanation:

Step1: Use angle - sum property of a triangle

The sum of the interior angles of a triangle is \(180^{\circ}\). So, \(x+3x+(3x + 5)=180\).

Step2: Combine like terms

\(x+3x+3x+5 = 180\), which simplifies to \(7x+5 = 180\).

Step3: Solve for \(x\)

Subtract 5 from both sides: \(7x=180 - 5=175\). Then divide both sides by 7: \(x=\frac{175}{7}=25\).

Step4: Find the angle measures

If \(x = 25\), then \(3x=3\times25 = 75\) and \(3x + 5=3\times25+5=75 + 5=80\).

Step5: Classify the triangle

Since all angles (\(25^{\circ}\), \(75^{\circ}\), \(80^{\circ}\)) are less than \(90^{\circ}\), it is an acute - angled triangle.

Answer:

The angle - measures are \(x = 25^{\circ}\), \(3x=75^{\circ}\), \(3x + 5=80^{\circ}\), and it is an acute - angled triangle.