Question

1. solve for the missing side lengths of the right triangles. a. b. c. d.

Explanation:

Step1: Use trigonometric ratio (a)

For right - triangle (a), we know the adjacent side \(a = 10\) to the angle \(\theta=31^{\circ}\) and we want to find the opposite side \(x\). Use the tangent ratio \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\).
\(\tan(31^{\circ})=\frac{x}{10}\)
\(x = 10\times\tan(31^{\circ})\)
Using a calculator, \(\tan(31^{\circ})\approx0.6009\), so \(x\approx10\times0.6009 = 6.009\approx6.0\)

Step2: Use trigonometric ratio (b)

For right - triangle (b), we know the hypotenuse \(c = 7\) and the angle \(\theta = 39^{\circ}\), and we want to find the adjacent side \(x\). Use the cosine ratio \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\)
\(\cos(39^{\circ})=\frac{x}{7}\)
\(x=7\times\cos(39^{\circ})\)
Using a calculator, \(\cos(39^{\circ})\approx0.7771\), so \(x\approx7\times0.7771 = 5.4397\approx5.4\)

Step3: Use trigonometric ratio (c)

For right - triangle (c), we know the opposite side \(a = 37\) to the angle \(\theta = 37^{\circ}\), and we want to find the hypotenuse \(x\). Use the sine ratio \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\)
\(\sin(37^{\circ})=\frac{37}{x}\)
\(x=\frac{37}{\sin(37^{\circ})}\)
Using a calculator, \(\sin(37^{\circ})\approx0.6018\), so \(x=\frac{37}{0.6018}\approx61.5\)

Step4: Use trigonometric ratio (d)

For right - triangle (d), we know the opposite side \(a = 20\) to the angle \(\theta = 38^{\circ}\), and we want to find the hypotenuse \(x\). Use the sine ratio \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\)
\(\sin(38^{\circ})=\frac{20}{x}\)
\(x=\frac{20}{\sin(38^{\circ})}\)
Using a calculator, \(\sin(38^{\circ})\approx0.6157\), so \(x=\frac{20}{0.6157}\approx32.5\)

Answer:

a. \(x\approx6.0\)
b. \(x\approx5.4\)
c. \(x\approx61.5\)
d. \(x\approx32.5\)