Sine Law And Cosine Law Word Problems

Introduction

The sine law and cosine law word problems are a staple of high‑school and college trigonometry courses because they bridge abstract formulas with real‑world situations. Whether you are calculating the height of a mountain, determining the distance between two ships, or figuring out the angle of a ramp, these laws let you solve any oblique triangle when you have enough information. Mastering them not only boosts your exam scores but also sharpens logical thinking and spatial reasoning—skills that prove useful in engineering, physics, architecture, and even everyday DIY projects. In this article we will break down the sine and cosine laws, outline a step‑by‑step strategy for tackling word problems, work through several detailed examples, explain the underlying derivations, and answer common questions that students often encounter.

Understanding the Sine Law

The law of sines relates the lengths of a triangle’s sides to the sines of its opposite angles. For any triangle with sides (a), (b), (c) and opposite angles (A), (B), (C),

[ \frac{a}{\sin A} = \frac{b}{\sin C} = \frac{c}{\sin C}. ]

Key points to remember:

• The ratio is constant for all three side‑angle pairs.
• It works best when you know either two angles and one side (AAS or ASA) or two sides and a non‑included angle (SSA).
• In the SSA case, the ambiguous case may produce zero, one, or two possible triangles; you must check the given data carefully.

When you see a word problem that mentions “angle of elevation,” “bearing,” or “distance across a river” and gives you two angles and a side, the sine law is usually the most direct route.

Understanding the Cosine Law

The law of cosines generalizes the Pythagorean theorem to any triangle. It is especially handy when you have two sides and the included angle (SAS) or all three sides (SSS). The formulas are:

[ \begin{aligned} a^{2} &= b^{2} + c^{2} - 2bc\cos A,\[4pt] b^{2} &= a^{2} + c^{2} - 2ac\cos B,\[4pt] c^{2} &= a^{2} + b^{2} - 2ab\cos C. \end{aligned} ]

Important notes:

• If the angle is (90^\circ), (\cos 90^\circ = 0) and the formula collapses to (a^{2}=b^{2}+c^{2}), the familiar Pythagorean theorem.
• The law of cosines eliminates the ambiguous case because the included angle uniquely determines the third side.
• After computing a side length, you can often revert to the sine law to find an unknown angle if needed.

Word problems that involve “distance between two points after traveling on different legs,” “resultant force,” or “length of a support beam” typically call for the cosine law.

Steps to Solve Sine and Cosine Law Word Problems

Follow this systematic approach to avoid missing details and to keep your work organized:

1. Read the problem carefully and identify what is being asked (a side length, an angle, a distance, a height, etc.).
2. Draw a diagram – even a rough sketch helps you label known sides and angles correctly. Mark the unknown quantity with a question mark.
3. List the given information in terms of sides ((a, b, c)) and angles ((A, B, C)). Use consistent units (convert if necessary).
4. Decide which law to apply:
   • Use the sine law if you have AAS, ASA, or the ambiguous SSA case.
   • Use the cosine law if you have SAS or SSS.
   • Sometimes you need both: first use cosine to find a side, then sine to find an angle.
5. Set up the appropriate equation and solve for the unknown. Keep track of whether you are solving for a side (take a square root at the end) or an angle (use the inverse sine or cosine function).
6. Check for ambiguity (only in SSA sine‑law situations). Compute the possible height (h = b\sin A) and compare it with the known side to see if zero, one, or two triangles exist.
7. Verify your answer by plugging it back into the original relationships or by using the triangle sum rule ((A+B+C=180^\circ)).
8. State the final answer with proper units and, if relevant, a brief interpretation (e.g., “The ladder reaches 7.3 m up the wall”).

Worked Example Problems

Example 1: Using the Sine Law (AAS)

Problem: From a point on the ground, the angle of elevation to the top of a hill is (28^\circ). After walking 150 m directly toward the hill, the angle of elevation becomes (45^\circ). Find the height of the hill.

Solution:

1. Sketch two right triangles sharing the hill’s vertical side (height (h)). Let the initial distance from the hill be (x); after walking 150 m, the remaining distance is (x-150).
2. In the first triangle: (\tan 28^\circ = \frac{h}{x}) → (h = x\tan 28^\circ).
3. In the second triangle: (\tan 45^\circ = \frac{h}{x-150}) → (h = (x-150)\tan 45^\circ = x-150) (since (\tan 45^\circ =1)).
4. Equate the two expressions for (h): (x\tan 28^\circ = x-150).
5. Solve for (x): (x(\tan 28^\circ -1) = -150) → (x = \frac{-150}{\tan 28^\circ -1}). Using (\tan 28^\circ \approx 0.5317), we get (x \approx \frac{-150}{-0.4683} \approx 320.3) m.
6. Compute (h = x-150 \approx 320.3-150 = 170.

… ≈ 170.Consider this: 3 m. Thus the hill’s height is about 170 m (rounded to the nearest metre) That alone is useful..

Example 2: Using the Cosine Law (SAS)

Problem: Two surveyors stand at points A and B on opposite sides of a river. From A they measure a distance of 120 m to a tree T, and from B they measure 95 m to the same tree. The angle ∠ATB between the two lines of sight is 62°. Find the width of the river, i.e., the distance AB.

Solution Sketch:

• Label triangle ATB with sides AT = 120 m, BT = 95 m, and included angle T = 62°.
• The unknown side AB is opposite angle T, so the cosine law applies directly.

Computation:
[ AB^{2}=AT^{2}+BT^{2}-2\cdot AT\cdot BT\cos T ] [ AB^{2}=120^{2}+95^{2}-2(120)(95)\cos 62^{\circ} ] [ AB^{2}=14400+9025-22800(0.4695)\approx 23425-10704\approx12721 ] [ AB\approx\sqrt{12721}\approx112.8\text{ m} ]

Check: Using the sine law to find one of the other angles (say A) gives
[ \frac{\sin A}{95}=\frac{\sin 62^{\circ}}{112.8};\Rightarrow;A\approx38.4^{\circ} ]
Then B = 180° − 62° − 38.4° ≈ 79.6°, and the sine law returns the original side lengths within rounding error, confirming the result Not complicated — just consistent..

Answer: The river is approximately 113 m wide.

Example 3: Ambiguous Case (SSA) with the Sine Law

Problem: From a ship S, the angle to a lighthouse L is 30°. The ship sails 200 m on a course that makes a 70° angle with the line SL, arriving at point P. At P the angle to the lighthouse is 45°. Determine the possible distances SL (from the ship’s original position to the lighthouse).

Interpretation: We have triangle SLP with known side SP = 200 m, known angle ∠S = 30° (at S), and known angle ∠P = 45° (at P). This is an ASA situation, so the sine law yields a unique side SL. Still, if the problem had given only one angle and two sides (SSA), we would need to check for ambiguity.

Solution (ASA):
First find the third angle:
[ \angle L = 180^{\circ} - 30^{\circ} - 45^{\circ}=105^{\circ} ]

Apply the sine law to find SL (opposite ∠P):
[ \frac{SL}{\sin 45^{\circ}} = \frac{200}{\sin 105^{\circ}} ] [ SL = 200;\frac{\sin 45^{\circ}}{\sin 105^{\circ}} = 200;\frac{0.7071}{0.9659} \approx 146 The details matter here. Surprisingly effective..

If the data were SSA (e.g., known SL = 150 m, SP = 200 m, ∠S = 30°), we would compute the height h = SP