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Question
problem 12: (first taught in lesson 30) find the measures of the angles of a right triangle where one of the two acute angles measures 8 times the other. after you enter your answer press go.
Step1: Let the smaller acute angle be \(x\)
Since one acute angle is 8 times the other, the larger acute angle is \(8x\). In a right - triangle, the sum of the two acute angles is \(90^{\circ}\) (because the sum of angles in a triangle is \(180^{\circ}\) and the right angle is \(90^{\circ}\)). So, we have the equation \(x + 8x=90\).
Step2: Solve the equation
Combine like terms: \(9x = 90\). Then, divide both sides by 9: \(x=\frac{90}{9}=10\).
Step3: Find the other acute angle
The larger acute angle is \(8x\). Substitute \(x = 10\) into \(8x\), we get \(8\times10 = 80\).
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\(10^{\circ},80^{\circ},90^{\circ}\)