Word problems for Law of Sines and Cosines with detailed solutions

📚 Introduction to the Laws of Sines and Cosines

The Law of Sines and the Law of Cosines are essential tools for solving triangles, especially when dealing with non-right triangles. These laws allow us to find unknown angles and sides using the relationships between angles and their opposite sides.

📜 History and Background

The Law of Sines and Cosines have ancient roots, with early forms appearing in the works of Greek mathematicians like Hipparchus and Ptolemy. These concepts were further developed by Islamic scholars during the Middle Ages and eventually made their way into modern trigonometry.

📐 Key Principles: Law of Sines

The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant for all three sides. Mathematically, it's expressed as:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

• 📏 a, b, c: Represent the lengths of the sides of the triangle.
• 📐 A, B, C: Represent the angles opposite those sides.

🔑 Key Principles: Law of Cosines

The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. There are three common forms:

• 💡 $a^2 = b^2 + c^2 - 2bc \cos A$
• 🔑 $b^2 = a^2 + c^2 - 2ac \cos B$
• 📌 $c^2 = a^2 + b^2 - 2ab \cos C$

🌍 Real-World Examples and Practice Problems

Problem 1: Surveying

A surveyor needs to determine the distance across a lake. From a point on one side of the lake, they measure the angle to a point on the other side to be 78 degrees. They then move 250 meters and measure the angle to the same point to be 42 degrees. Find the distance across the lake.

Solution:

1. Let $d$ be the distance across the lake.
2. Using the Law of Sines: $\frac{d}{\sin 42^\circ} = \frac{250}{\sin (180^\circ - 78^\circ - 42^\circ)}$ $\frac{d}{\sin 42^\circ} = \frac{250}{\sin 60^\circ}$ $d = \frac{250 \cdot \sin 42^\circ}{\sin 60^\circ} \approx 189.3$ meters

Problem 2: Navigation

A ship sails 40 miles east and then turns 30 degrees north of east and sails another 30 miles. How far is the ship from its starting point?

Solution:

1. Let $d$ be the distance from the starting point.
2. Using the Law of Cosines: $d^2 = 40^2 + 30^2 - 2 \cdot 40 \cdot 30 \cdot \cos 150^\circ$ $d^2 = 1600 + 900 - 2400 \cdot (-\frac{\sqrt{3}}{2})$ $d^2 = 2500 + 1200\sqrt{3}$ $d = \sqrt{2500 + 1200\sqrt{3}} \approx 67.67$ miles

Problem 3: Triangle Properties

In triangle ABC, angle A = 55 degrees, side b = 15, and side a = 12. Find angle B.

Solution:

1. Using the Law of Sines: $\frac{\sin B}{b} = \frac{\sin A}{a}$ $\frac{\sin B}{15} = \frac{\sin 55^\circ}{12}$ $\sin B = \frac{15 \cdot \sin 55^\circ}{12} \approx 1.0239$ Since the sine value cannot be greater than 1, there is no solution for angle B in this case. This indicates that the given triangle is impossible.

Problem 4: Height of a Tree

To find the height of a tree, a surveyor measures the angle of elevation to the top of the tree to be 35 degrees. They then move back 20 feet and measure the angle of elevation to be 20 degrees. How tall is the tree?

Solution:

1. Let $h$ be the height of the tree, and $x$ be the initial distance from the tree.
2. $\frac{h}{\sin 20^\circ} = \frac{20}{\sin (35^\circ - 20^\circ)}$ $h = \frac{20 \cdot \sin 20^\circ}{\sin 15^\circ} \approx 26.4$ feet

Problem 5: Bridge Length

A bridge is to be built across a river. The distance from point A to point C is measured to be 350 feet. Angle BAC is 82 degrees and angle ACB is 45 degrees. How long must the bridge be?

Solution:

1. Let $b$ be the length of the bridge.
2. Using the Law of Sines: $\frac{b}{\sin 82^\circ} = \frac{350}{\sin (180^\circ - 82^\circ - 45^\circ)}$ $\frac{b}{\sin 82^\circ} = \frac{350}{\sin 53^\circ}$ $b = \frac{350 \cdot \sin 82^\circ}{\sin 53^\circ} \approx 436.9$ feet

Problem 6: Airplane Distance

Two observers 5 miles apart see an airplane. The angle of elevation to the airplane is 35 degrees from one observer and 65 degrees from the other. How far is the airplane from each observer?

Solution:

1. Let $d_1$ and $d_2$ be the distances from the observers to the airplane.
2. Using the Law of Sines: $\frac{d_1}{\sin 65^\circ} = \frac{5}{\sin (180^\circ - 35^\circ - 65^\circ)}$ $\frac{d_1}{\sin 65^\circ} = \frac{5}{\sin 80^\circ}$ $d_1 = \frac{5 \cdot \sin 65^\circ}{\sin 80^\circ} \approx 4.6$ miles $\frac{d_2}{\sin 35^\circ} = \frac{5}{\sin 80^\circ}$ $d_2 = \frac{5 \cdot \sin 35^\circ}{\sin 80^\circ} \approx 2.9$ miles

Problem 7: Tower Height

A tower is leaning. A cable 40 feet long is attached to the top of the tower and makes an angle of 63 degrees with the ground. The distance from the base of the tower to the point where the cable is anchored is 28 feet. How far is the top of the tower from the ground?

Solution:

1. Let $h$ be the height of the tower.
2. Using the Law of Cosines: $40^2 = 28^2 + h^2 - 2 \cdot 28 \cdot h \cdot \cos 63^\circ$ $1600 = 784 + h^2 - 56h \cdot \cos 63^\circ$ $h^2 - 56 \cos 63^\circ \cdot h - 816 = 0$ $h^2 - 25.43h - 816 = 0$ Using the quadratic formula, $h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $h = \frac{25.43 \pm \sqrt{(-25.43)^2 - 4(1)(-816)}}{2(1)}$ $h \approx 41.7$ feet

🔑 Conclusion

The Law of Sines and the Law of Cosines are powerful tools for solving various problems involving triangles. Understanding these laws and practicing with real-world examples can greatly enhance your problem-solving skills in trigonometry.