The Cosine Rule: It’s easier than it seems

The Cosine Rule

It’s Easier Than It Seems

GCSE Trigonometry

Main Formula

\[ a^2=b^2+c^2-2bc\cos(A) \]

The Cosine Rule allows us to solve triangles that do not contain a right angle.

When students first see the Cosine Rule, many think the formula looks difficult.

But the process is actually very repetitive and structured.

Every question follows the same idea:

\[ \text{identify the sides} \rightarrow \text{substitute carefully} \rightarrow \text{solve step-by-step} \]

The biggest challenge is usually matching the correct angle with the correct opposite side.

Side \(a\) is opposite angle \(A\)

1. When should we use the Cosine Rule?

The Cosine Rule is used in non-right-angled triangles.

It is especially useful in two situations:

1. when we know two sides and the angle between them;
2. when we know all three sides and need an angle.

Quick memory rule

If SOHCAHTOA does not work because there is no right angle, the Cosine Rule is often the next method to try.

2. Understanding the Formula

The Cosine Rule is:

\[ a^2=b^2+c^2-2bc\cos(A) \]

At first glance, this formula looks complicated.

But the structure is actually simple.

\[ \text{side squared} = \text{side squared} + \text{side squared} - \text{adjustment} \]

The cosine part changes the answer depending on the size of the angle.

• Small angles create shorter opposite sides.
• Large angles create longer opposite sides.

3. Example — Finding a Missing Side

Question

Find the missing side \(x\).

Step 1 — Identify what we know

We know:

\[ 5\text{ cm}, \quad 8\text{ cm}, \quad 60^\circ \]

The angle is between the two known sides.

That means this is a Cosine Rule question.

We are trying to find the side opposite the angle.

So:

\[ a=x \]

Step 2 — Write the formula

\[ a^2=b^2+c^2-2bc\cos(A) \]

Step 3 — Substitute values carefully

Replace each letter with the correct number.

\[ x^2=5^2+8^2-2(5)(8)\cos(60^\circ) \]

Step 4 — Calculate powers first

\[ x^2=25+64-80\cos(60^\circ) \]

Now evaluate the cosine.

\[ \cos(60^\circ)=0.5 \]

So:

\[ x^2=25+64-80(0.5) \]

Step 5 — Simplify

\[ x^2=89-40 \]

\[ x^2=49 \]

Step 6 — Square root both sides

We need \(x\), not \(x^2\).

So we square root both sides.

\[ x=\sqrt{49} \]

\[ x=7 \]

\[ \boxed{x=7\text{ cm}} \]

Always remember:

If the formula gives \(x^2\), the final step is square rooting.

4. Example — Finding a Missing Angle

Question

Find angle \(A\).

Step 1 — Identify the opposite side

The angle we need is \(A\).

The side opposite angle \(A\) is:

\[ 11\text{ cm} \]

So:

\[ a=11 \]

The remaining sides are:

\[ b=6,\quad c=9 \]

Step 2 — Rearranged formula

When finding an angle, it is easier to use:

\[ \cos(A)=\frac{b^2+c^2-a^2}{2bc} \]

Step 3 — Substitute values

\[ \cos(A)=\frac{6^2+9^2-11^2}{2(6)(9)} \]

Step 4 — Evaluate powers

\[ \cos(A)=\frac{36+81-121}{108} \]

\[ \cos(A)=\frac{-4}{108} \]

\[ \cos(A)\approx -0.037 \]

Step 5 — Use inverse cosine

At the moment we have the cosine value.

But we need the angle itself.

So we use inverse cosine.

\[ A=\cos^{-1}(-0.037) \]

\[ A\approx92^\circ \]

\[ \boxed{A\approx92^\circ} \]

When finding an angle:

\[ \cos^{-1} \]

is one of the most important calculator buttons.

5. Exercises

Exercise 1

Find the missing side \(x\).

Solution

We know two sides and the angle between them.

That means we use the Cosine Rule.

The side we are finding is opposite the angle.

So:

\[ a=x,\quad b=7,\quad c=10 \]

Write the formula:

\[ a^2=b^2+c^2-2bc\cos(A) \]

Substitute:

\[ x^2=7^2+10^2-2(7)(10)\cos(45^\circ) \]

Calculate powers:

\[ x^2=49+100-140\cos(45^\circ) \]

Now evaluate the cosine:

\[ \cos(45^\circ)\approx0.7071 \]

Substitute:

\[ x^2=149-140(0.7071) \]

\[ x^2=149-98.99 \]

\[ x^2\approx50.01 \]

Square root both sides:

\[ x\approx\sqrt{50.01} \]

\[ x\approx7.1 \]

\[ \boxed{x\approx7.1\text{ cm}} \]

Exercise 2

Find the missing side \(x\).

Solution

We know:

\[ 6\text{ cm}, \quad 9\text{ cm}, \quad 110^\circ \]

The angle is between the known sides.

So we use the Cosine Rule.

\[ x^2=6^2+9^2-2(6)(9)\cos(110^\circ) \]

Calculate powers:

\[ x^2=36+81-108\cos(110^\circ) \]

Now evaluate the cosine.

\[ \cos(110^\circ)\approx -0.342 \]

Notice that the cosine is negative because the angle is obtuse.

Substitute:

\[ x^2=117-108(-0.342) \]

Subtracting a negative becomes addition.

\[ x^2=117+36.94 \]

\[ x^2\approx153.94 \]

Now square root both sides:

\[ x\approx\sqrt{153.94} \]

\[ x\approx12.4 \]

\[ \boxed{x\approx12.4\text{ cm}} \]

Exercise 3

Find angle \(A\).

Solution

All three sides are known.

That means we use the Cosine Rule to find an angle.

The side opposite angle \(A\) is:

\[ 7\text{ cm} \]

So:

\[ a=7,\quad b=8,\quad c=10 \]

Use the rearranged formula:

\[ \cos(A)=\frac{b^2+c^2-a^2}{2bc} \]

Substitute values:

\[ \cos(A)=\frac{8^2+10^2-7^2}{2(8)(10)} \]

Evaluate powers:

\[ \cos(A)=\frac{64+100-49}{160} \]

\[ \cos(A)=\frac{115}{160} \]

\[ \cos(A)=0.71875 \]

Now use inverse cosine:

\[ A=\cos^{-1}(0.71875) \]

\[ A\approx44^\circ \]

\[ \boxed{A\approx44^\circ} \]

Exercise 4

A triangular garden has sides of length \(12\) m and \(15\) m.

The angle between them is \(70^\circ\).

Find the third side.

Solution

This is a real-life problem, but mathematically it is still a Cosine Rule question.

We know two sides and the included angle.

Write the formula:

\[ x^2=12^2+15^2-2(12)(15)\cos(70^\circ) \]

Calculate powers:

\[ x^2=144+225-360\cos(70^\circ) \]

Evaluate the cosine:

\[ \cos(70^\circ)\approx0.342 \]

Substitute:

\[ x^2=369-360(0.342) \]

\[ x^2=369-123.12 \]

\[ x^2\approx245.88 \]

Square root:

\[ x\approx\sqrt{245.88} \]

\[ x\approx15.7 \]

\[ \boxed{x\approx15.7\text{ m}} \]

Exercise 5

A triangle has side lengths:

\[ 9\text{ cm},\quad 11\text{ cm},\quad 14\text{ cm} \]

Find the angle opposite the side of length \(14\) cm.

Solution

The angle we need is opposite the side of length \(14\) cm.

So:

\[ a=14,\quad b=9,\quad c=11 \]

Use the rearranged formula:

\[ \cos(A)=\frac{b^2+c^2-a^2}{2bc} \]

Substitute:

\[ \cos(A)=\frac{9^2+11^2-14^2}{2(9)(11)} \]

Evaluate powers:

\[ \cos(A)=\frac{81+121-196}{198} \]

\[ \cos(A)=\frac{6}{198} \]

\[ \cos(A)\approx0.0303 \]

Now use inverse cosine:

\[ A=\cos^{-1}(0.0303) \]

\[ A\approx88^\circ \]

\[ \boxed{A\approx88^\circ} \]

Because the cosine value is very close to zero, the angle is very close to \(90^\circ\).

Exercise 6 — Real Life Application

A rescue team is planning a zipline across a canyon.

One side of the canyon is connected to a point that is \(120\) m away from the launch platform.

Another support point is \(85\) m away from the same platform.

The angle between the two support cables is \(48^\circ\).

Find the distance between the two support points.

Give your answer to 1 decimal place.

Solution

This is a real-life problem involving distances and an angle.

We know:

\[ 120\text{ m}, \quad 85\text{ m}, \quad 48^\circ \]

The angle is between the two known sides.

That means this is a Cosine Rule question.

We are trying to find the side opposite the angle.

So:

\[ a=x,\quad b=120,\quad c=85 \]

Write the Cosine Rule:

\[ a^2=b^2+c^2-2bc\cos(A) \]

Substitute the values:

\[ x^2=120^2+85^2-2(120)(85)\cos(48^\circ) \]

Calculate the powers first:

\[ x^2=14400+7225-20400\cos(48^\circ) \]

Now evaluate the cosine:

\[ \cos(48^\circ)\approx0.6691 \]

Substitute this value:

\[ x^2=21625-20400(0.6691) \]

Multiply carefully:

\[ x^2=21625-13649.64 \]

\[ x^2\approx7975.36 \]

Now square root both sides to find \(x\):

\[ x\approx\sqrt{7975.36} \]

\[ x\approx89.3 \]

\[ \boxed{x\approx89.3\text{ m}} \]

Notice how this real-life problem still follows exactly the same steps:

\[ \text{identify the sides} \rightarrow \text{substitute carefully} \rightarrow \text{calculate step-by-step} \]

The context changes, but the mathematics stays the same.

6. Final Summary

1. Identify the opposite side carefully.
2. Match the angle with the correct side letter.
3. Substitute slowly and carefully.
4. Use brackets properly.
5. Use \(\cos^{-1}\) for missing angles.
6. Square root at the end for missing sides.

\[ \boxed{ \text{match the sides} \rightarrow \text{substitute carefully} \rightarrow \text{solve step-by-step} } \]

Final Checklist

• Is the calculator in degree mode?
• Did you identify the opposite side correctly?
• Did you use inverse cosine for angles?
• Did you square root at the end for sides?