A ship is sailing due north. At a certain point the bearing of a lighthouse 12.5 km away is N 38.8° E. Later on, the captain notices that the bearing of the lighthouse has become S 44.2° E. How far did the ship travel between the two observations of the lighthouse?
Table of contents
- 0. Review of College Algebra4h 45m
- 1. Measuring Angles40m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
7. Non-Right Triangles
Law of Sines
Multiple Choice
Use the Law of Sines to find the angle B to the nearest tenth of a degree.

A
48.6°
B
77.2°
C
40.5°
D
35.3°
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Verified step by step guidance1
Identify the given values in the triangle: side a = 4, side b = 6, side c = 7.8, and angle C = 30°.
Recall the Law of Sines formula: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \).
Use the Law of Sines to set up the equation for angle B: \( \frac{b}{\sin B} = \frac{c}{\sin C} \). Substitute the known values: \( \frac{6}{\sin B} = \frac{7.8}{\sin 30°} \).
Calculate \( \sin 30° \), which is 0.5, and substitute it into the equation: \( \frac{6}{\sin B} = \frac{7.8}{0.5} \).
Solve for \( \sin B \) by cross-multiplying and simplifying: \( \sin B = \frac{6 \times 0.5}{7.8} \). Then, use the inverse sine function to find angle B to the nearest tenth of a degree.
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