How to Solve: Law of Cosines

How to Solve: Law of Cosines

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

Introduction

"Ever tried to find the length of a side in a triangle when you only know two sides and the included angle? The Law of Cosines is your secret weapon—it’s how engineers design bridges, pilots calculate flight paths, and you’ll solve those tricky exam questions in under 60 seconds!

What You Need To Know First

Before diving into the Law of Cosines, make sure you understand: 1. Pythagorean Theorem – Only works for right triangles. The Law of Cosines is its "big brother" for any triangle. 2. Basic Trigonometry (SOH-CAH-TOA) – You’ll need to identify sides (opposite, adjacent, hypotenuse) and angles. 3. Angle-Side Relationships – In any triangle, the largest angle is opposite the longest side, and vice versa.

If any of these feel shaky, pause and review them first!

Key Vocabulary

Formulas To Know

1. Law of Cosines (for finding a side)

Formula: [ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) ]

Variables: - a, b, c = lengths of the sides of the triangle. - C = the angle opposite side c (the side you’re solving for).

When to use: - SAS (Side-Angle-Side): You know two sides and the included angle. - SSS (Side-Side-Side): You know all three sides and need to find an angle (see next formula).

MEMORISE THIS? ✅ Yes! (But it’s often given on exam sheets—check your formula booklet.)

2. Law of Cosines (for finding an angle)

Formula: [ \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} ]

Variables: - Same as above, but now you’re solving for angle C.

When to use: - SSS (Side-Side-Side): You know all three sides and need to find an angle.

MEMORISE THIS? ❌ No—just rearrange the first formula.

3. Pythagorean Theorem (Special Case)

Formula: [ c^2 = a^2 + b^2 ]

When to use: - Only for right triangles (angle C = 90°). The Law of Cosines becomes the Pythagorean Theorem when cos(90°) = 0.

MEMORISE THIS? ✅ Yes (but it’s a subset of the Law of Cosines).

Step-by-Step Method

How to Solve Using the Law of Cosines

Step 1: Draw the Triangle - Sketch the triangle and label all known sides and angles. - Circle the unknown you’re solving for (side or angle).

Step 2: Identify the Scenario - SAS? Use the side formula: ( c^2 = a^2 + b^2 - 2ab \cdot \cos(C) ). - SSS? Use the angle formula: ( \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} ).

Step 3: Plug in the Values - Substitute the known values into the correct formula. - Double-check: Is the angle opposite the side you’re solving for?

Step 4: Solve for the Unknown - For sides: Take the square root at the end. - For angles: Use ( \cos^{-1} ) (inverse cosine) to find the angle.

Step 5: Round and Check Reasonableness - Round to the required decimal places (usually 1 or 2). - Ask: Does the answer make sense? (e.g., largest angle opposite longest side).

Worked Example (Using the Steps)

Problem: In △ABC, a = 7 cm, b = 10 cm, and angle C = 45°. Find side c.

Step 1: Draw the Triangle

       C

       

      / \
  7 /   \ 10

    /     \
A ------- B

      c

• Known: a = 7, b = 10, C = 45°.
• Unknown: c.

Step 2: Identify the Scenario - SAS (two sides and the included angle). - Use: ( c^2 = a^2 + b^2 - 2ab \cdot \cos(C) ).

Step 3: Plug in the Values [ c^2 = 7^2 + 10^2 - 2(7)(10) \cdot \cos(45°) ] [ c^2 = 49 + 100 - 140 \cdot \cos(45°) ]

Step 4: Solve for c - ( \cos(45°) = \frac{\sqrt{2}}{2} \approx 0.7071 ). - ( c^2 = 149 - 140(0.7071) ). - ( c^2 = 149 - 98.994 ). - ( c^2 = 50.006 ). - ( c = \sqrt{50.006} \approx 7.07 ) cm.

Step 5: Round and Check - Rounded to 2 decimal places: c ≈ 7.07 cm. - Check: c is between a and b (7 < 7.07 < 10), which makes sense.

Worked Examples

Example 1 – Basic (SAS)

Problem: In △PQR, p = 5 m, q = 8 m, and angle R = 60°. Find r.

Solution: 1. SAS → Use ( r^2 = p^2 + q^2 - 2pq \cdot \cos(R) ). 2. Plug in: ( r^2 = 5^2 + 8^2 - 2(5)(8) \cdot \cos(60°) ). 3. ( \cos(60°) = 0.5 ). 4. ( r^2 = 25 + 64 - 80(0.5) = 89 - 40 = 49 ). 5. ( r = \sqrt{49} = 7 ) m.

What we did and why: - We used the Law of Cosines because we had two sides and the included angle (SAS). - The calculation simplified nicely because ( \cos(60°) ) is a clean fraction.

Example 2 – Medium (SSS to Find Angle)

Problem: In △XYZ, x = 9 cm, y = 11 cm, z = 7 cm. Find angle X.

Solution: 1. SSS → Use ( \cos(X) = \frac{y^2 + z^2 - x^2}{2yz} ). 2. Plug in: ( \cos(X) = \frac{11^2 + 7^2 - 9^2}{2(11)(7)} ). 3. ( \cos(X) = \frac{121 + 49 - 81}{154} = \frac{89}{154} \approx 0.5779 ). 4. ( X = \cos^{-1}(0.5779) \approx 54.7° ).

What we did and why: - We rearranged the Law of Cosines to solve for an angle because all three sides were known (SSS). - Always check: The largest angle (X ≈ 54.7°) is opposite the longest side (y = 11 cm).

Example 3 – Exam Style (Disguised Problem)

Problem: A ship leaves port and travels 12 km north, then turns 30° east and travels another 8 km. How far is the ship from the port? (Round to 1 decimal place.)

Solution: 1. Sketch the scenario:
- Port at point A.
- First leg: 12 km north to point B.
- Second leg: 8 km at 30° east of north to point C.
- We need distance AC.

1. Identify the triangle:
2. △ABC with AB = 12 km, BC = 8 km, and angle B = 150° (180° - 30°).

3. SAS → Use Law of Cosines.

4. Apply the formula:
   [ AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cdot \cos(B) ]
   [ AC^2 = 12^2 + 8^2 - 2(12)(8) \cdot \cos(150°) ]
   [ \cos(150°) = -\cos(30°) = -\frac{\sqrt{3}}{2} \approx -0.8660 ]
   [ AC^2 = 144 + 64 - 192(-0.8660) ]
   [ AC^2 = 208 + 166.272 = 374.272 ]
   [ AC = \sqrt{374.272} \approx 19.3 \text{ km} ]

What we did and why: - We recognized the problem as a triangle with two sides and the included angle (SAS). - The angle was disguised (150° instead of 30°), so we had to adjust for the turn. - The negative cosine was crucial—missing it would give the wrong answer!

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in. The Law of Cosines is your go-to when you’ve got: - Two sides and the included angle (SAS), or - All three sides (SSS) and need an angle.

Here’s how to crush it: 1. Label your triangle—side a opposite angle A, side b opposite angle B, etc. 2. Pick the right formula:
- For a side: ( c^2 = a^2 + b^2 - 2ab \cdot \cos(C) ).
- For an angle: ( \cos(C) = \frac{a^2 + b^2 - c^2}{2ab} ). 3. Plug in the numbers—double-check the angle is between the two sides. 4. Solve and round—don’t forget the square root for sides or ( \cos^{-1} ) for angles. 5. Check your answer—does the largest angle face the longest side? If not, redo it!

Pro tip: If the problem mentions a ‘turn’ or ‘bearing,’ sketch it first—the angle you need might not be the one given!

You’ve got this. Now go ace that exam!