Question

math 3: starter week 8 friday (9/26/25) finding the side or angle of a right triangle

1. identify the reference angle
2. label the sides
3. identify the trig ratios
4. substitute
5. solve!

page 5 / 5

Explanation:

Answer:

For the first triangle:

1. Reference angle: \(60^{\circ}\)
2. Labeling sides: Let the side opposite \(\angle B\) be \(x\), the side adjacent to \(\angle B\) be the side of length \(11\), and the hypotenuse be \(AB\).
3. Trig - ratios: \(\sin B=\frac{x}{AB}\), \(\cos B = \frac{11}{AB}\), \(\tan B=\frac{x}{11}\)
4. Substitute: Since \(\tan B=\tan60^{\circ}=\sqrt{3}\) and \(\tan B=\frac{x}{11}\), we substitute \(\sqrt{3}=\frac{x}{11}\)
5. Solve: \(x = 11\sqrt{3}\)

For the second triangle:

1. Reference angle: \(37^{\circ}\)
2. Labeling sides: Let the side opposite \(\angle B\) be \(AC\), the side adjacent to \(\angle B\) be \(BC\), and the hypotenuse be \(AB = 10.3\)
3. Trig - ratios: \(\sin B=\frac{AC}{AB}\), \(\cos B=\frac{BC}{AB}\), \(\tan B=\frac{AC}{BC}\)
4. Substitute: If we want to find \(AC\), since \(\sin B=\sin37^{\circ}\approx0.6018\) and \(\sin B=\frac{AC}{10.3}\), we substitute \(0.6018=\frac{AC}{10.3}\)
5. Solve: \(AC\approx10.3\times0.6018 = 6.2\)