How to Solve: Law of Sines

How to Solve: Law of Sines

(For Students Who Want to Ace Their Exam & Teachers Who Need a Ready-to-Record Script)

Introduction

"Ever wondered how surveyors measure the height of a mountain without climbing it? Or how pilots calculate their flight path when winds change? The Law of Sines is the secret tool—and it’s guaranteed to appear on your trigonometry exam. Master it today, and you’ll solve triangles like a pro in under 60 seconds."

What You Need To Know First

Before diving into the Law of Sines, ensure you understand:
1. Basic trigonometric ratios (sine, cosine, tangent) – You must know how to find sine of an angle in a right triangle.
2. Parts of a triangle (angles and sides) – Labeling sides a, b, c opposite angles A, B, C is critical.
3. Degrees vs. radians – Most exam questions use degrees, but check the unit!

Key Vocabulary

Formulas To Know

1. Law of Sines (MEMORISE THIS)

Formula: [ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} ] What it means: - a, b, c = lengths of sides opposite angles A, B, C. - The ratio of any side to the sine of its opposite angle is the same for all three sides in a triangle.

When to use it: - ASA (Angle-Side-Angle) – Two angles and the included side. - AAS (Angle-Angle-Side) – Two angles and a non-included side. - SSA (Side-Side-Angle) – Two sides and a non-included angle (ambiguous case!).

2. Area of a Triangle (Given on Exam Sheet, but Useful to Know)

Formula: [ \text{Area} = \frac{1}{2}ab \sin C ] What it means: - a and b = two sides of the triangle. - C = the included angle between sides a and b.

When to use it: - When you know two sides and the included angle (SAS).

Step-by-Step Method

How to Solve Any Law of Sines Problem

Step 1: Label the triangle - Write down all given information (angles and sides). - Label sides a, b, c opposite angles A, B, C.

Step 2: Check if the Law of Sines applies - Yes if you have: - Two angles and one side (ASA or AAS). - Two sides and a non-included angle (SSA – watch for ambiguity!). - No if you have: - Three sides (use Law of Cosines). - Two sides and the included angle (use Law of Cosines or Area formula).

Step 3: Write the Law of Sines ratio - Pick two fractions from the formula that include one known side and its opposite angle. - Example: If you know a and ∠A, and need b or ∠B, write: [ \frac{a}{\sin A} = \frac{b}{\sin B} ]

Step 4: Solve for the unknown - Cross-multiply and solve for the missing variable. - If solving for an angle, use inverse sine (sin⁻¹).

Step 5: Check for the ambiguous case (SSA only!) - If given two sides and a non-included angle (SSA), calculate: [ h = b \sin A ] - If a < h → No triangle exists. - If a = h → One right triangle exists. - If h < a < b → Two different triangles exist. - If a ≥ b → One triangle exists.

Step 6: Find all missing parts - If you found an angle, subtract from 180° to find the third angle. - Use the Law of Sines again to find the remaining side(s).

Step 7: Verify your answer - Check that the sum of angles = 180°. - Ensure side lengths make sense (longest side opposite largest angle).

Worked Example (Using Exact Steps)

Problem: In triangle ABC, ∠A = 40°, ∠B = 60°, and a = 5 cm. Find side b.

Step 1: Label the triangle - ∠A = 40°, ∠B = 60°, a = 5 cm (opposite ∠A). - Need to find b (opposite ∠B).

Step 2: Check if Law of Sines applies - We have two angles and one side (AAS) → Yes.

Step 3: Write the Law of Sines ratio [ \frac{a}{\sin A} = \frac{b}{\sin B} ] [ \frac{5}{\sin 40°} = \frac{b}{\sin 60°} ]

Step 4: Solve for b - Cross-multiply: [ b \sin 40° = 5 \sin 60° ] - Isolate b: [ b = \frac{5 \sin 60°}{\sin 40°} ] - Calculate (use calculator in degree mode!): [ \sin 60° ≈ 0.8660, \quad \sin 40° ≈ 0.6428 ] [ b ≈ \frac{5 \times 0.8660}{0.6428} ≈ \frac{4.33}{0.6428} ≈ 6.74 \text{ cm} ]

Step 5: Check for ambiguity (not SSA, so skip)

Step 6: Find all missing parts (optional here, but good practice) - ∠C = 180° – 40° – 60° = 80°. - Find c using Law of Sines: [ \frac{c}{\sin 80°} = \frac{5}{\sin 40°} \implies c ≈ \frac{5 \times 0.9848}{0.6428} ≈ 7.66 \text{ cm} ]

Step 7: Verify - Angles: 40° + 60° + 80° = 180° ✔️ - Sides: a = 5 (smallest), b ≈ 6.74, c ≈ 7.66 (largest) → matches angles ✔️

Final Answer: [ b ≈ 6.74 \text{ cm} ]

What we did and why: - We used AAS (two angles and a side), so the Law of Sines was perfect. - We set up the ratio with known side a and its opposite angle A, then solved for b. - Always check angle sums to avoid mistakes!

Worked Examples

Example 1 – Basic (AAS)

Problem: In triangle DEF, ∠D = 50°, ∠E = 70°, and d = 10 m. Find side e.

Solution:
1. Label: ∠D = 50°, ∠E = 70°, d = 10 m (opposite ∠D).
2. Law of Sines applies (AAS).
3. Write ratio: [ \frac{d}{\sin D} = \frac{e}{\sin E} \implies \frac{10}{\sin 50°} = \frac{e}{\sin 70°} ]
4. Solve for e: [ e = \frac{10 \sin 70°}{\sin 50°} ≈ \frac{10 \times 0.9397}{0.7660} ≈ 12.27 \text{ m} ]
5. Verify: ∠F = 60°, angles sum to 180° ✔️

Answer: [ e ≈ 12.3 \text{ m} ]

What we did and why: - AAS is straightforward—just plug into the Law of Sines. - Always use the side opposite the known angle in the ratio.

Example 2 – Medium (SSA – Ambiguous Case)

Problem: In triangle PQR, p = 8 cm, q = 6 cm, and ∠P = 35°. Find ∠Q.

Solution:
1. Label: p = 8 cm (opposite ∠P), q = 6 cm (opposite ∠Q), ∠P = 35°.
2. Law of Sines applies (SSA) → check for ambiguity.
3. Calculate h: [ h = q \sin P = 6 \sin 35° ≈ 6 \times 0.5736 ≈ 3.44 \text{ cm} ]
4. Compare p and h: - p = 8 cm, h ≈ 3.44 cm → h < p < q → Two possible triangles.
5. Use Law of Sines to find ∠Q: [ \frac{p}{\sin P} = \frac{q}{\sin Q} \implies \frac{8}{\sin 35°} = \frac{6}{\sin Q} ] [ \sin Q = \frac{6 \sin 35°}{8} ≈ \frac{6 \times 0.5736}{8} ≈ 0.4302 ] [ Q ≈ \sin^{-1}(0.4302) ≈ 25.5° \quad \text{or} \quad Q ≈ 180° - 25.5° = 154.5° ]
6. Check validity: - If Q ≈ 25.5°, then ∠R ≈ 119.5° (valid). - If Q ≈ 154.5°, then ∠R ≈ -9.5° (invalid, angles can’t be negative). - Only one valid triangle: Q ≈ 25.5°.

Answer: [ ∠Q ≈ 25.5° ]

What we did and why: - SSA is tricky—always check for two possible triangles. - If sin⁻¹ gives an angle, subtract from 180° to check for a second solution. - Discard invalid angles (sum > 180° or negative).

Example 3 – Exam Style (Disguised Problem)

Problem: A surveyor measures the angle of elevation to the top of a building as 28° from point A. She walks 50 m toward the building to point B and measures the angle of elevation as 42°. Find the height of the building.

Solution:
1. Draw the diagram: - Points A and B on the ground, building = point C. - ∠CAB = 28°, ∠CBA = 180° – 42° = 138° (supplementary angles). - AB = 50 m.
2. Find ∠ACB: [ ∠ACB = 180° – 28° – 138° = 14° ]
3. Use Law of Sines to find AC: [ \frac{AC}{\sin 138°} = \frac{AB}{\sin 14°} \implies \frac{AC}{\sin 42°} = \frac{50}{\sin 14°} ] (Note: sin(138°) = sin(42°)) [ AC = \frac{50 \sin 42°}{\sin 14°} ≈ \frac{50 \times 0.6691}{0.2419} ≈ 138.3 \text{ m} ]
4. Find height (h) using right triangle ACD (D = base of building): [ \sin 28° = \frac{h}{AC} \implies h = AC \sin 28° ≈ 138.3 \times 0.4695 ≈ 65.0 \text{ m} ]

Answer: [ \text{Height} ≈ 65.0 \text{ m} ]

What we did and why: - Real-world problems often hide the triangle—draw a diagram first. - Supplementary angles (180° – 42° = 138°) are key. - Two steps: (1) Find a side using Law of Sines, (2) Use sine in a right triangle for height.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in—tonight, before your exam. Here’s the Law of Sines in 60 seconds:

1. When to use it: Two angles and a side (ASA/AAS) or two sides and a non-included angle (SSA—watch for ambiguity!).
2. The formula: [ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} ]
3. Side a is opposite angle A. Don’t mix them up!
4. Steps:
5. Label the triangle.
6. Write the ratio with one known side-angle pair.
7. Cross-multiply and solve.
8. For SSA, check h = b sin A—if a < h, no triangle. If h < a < b, two triangles!
9. Common traps:
10. SSA ambiguity—always check for two solutions.
11. Wrong mode—calculator in degrees, not radians.
12. Missing the third angle—always find it to verify.
13. Exam tip: If you see two angles adding to 90°, it’s probably a right triangle—don’t use Law of Sines!

You’ve got this. Now go practice two problems—one AAS, one SSA—and you’ll own this topic. Good luck!