Question

problem 15: (first taught in lesson 30) find the measures of the angles of a right triangle where one of the two acute angles measures 5 times the other. after you enter your answer press go. go

Explanation:

Step1: Let the smaller acute angle be \(x\)

In a right - triangle, the sum of the two acute angles is \(90^{\circ}\). Let the smaller acute angle be \(x\), then the larger acute angle is \(5x\).

Step2: Set up the equation

We know that \(x + 5x=90^{\circ}\).
Simplify the left - hand side: \(6x = 90^{\circ}\).

Step3: Solve for \(x\)

Divide both sides of the equation \(6x = 90^{\circ}\) by \(6\): \(x=\frac{90^{\circ}}{6}=15^{\circ}\).
The larger acute angle is \(5x = 5\times15^{\circ}=75^{\circ}\), and the right - angle is \(90^{\circ}\).

Answer:

\(15^{\circ},75^{\circ},90^{\circ}\)