Lesson 10.2 : Non-right Triangles: Law of Cosines

New Concept

Master the Law of Cosines, a key tool for solving oblique (non-right) triangles in cases where the Law of Sines falls short, like SAS and SSS. You'll learn to find unknown sides, angles, and calculate area using Heron's formula.

What’s next

Next, you'll dive into worked examples for SAS and SSS triangles. Then, test your new skills with a series of interactive practice cards.

Property

The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles with angles $\alpha$, $\beta$, and $\gamma$, and opposite corresponding sides $a$, $b$, and $c$, the Law of Cosines is given as three equations.

To solve for an angle, these can be rearranged, for example:

Examples

• To find side $c$ in a triangle where $a=8$, $b=10$, and $\gamma=60^\circ$: $c^2 = 8^2 + 10^2 - 2(8)(10)\operatorname{cos}(60^\circ) = 64 + 100 - 160(0.5) = 84$. So, $c = \sqrt{84} \approx 9.17$.
• To find angle $\alpha$ in a triangle with sides $a=7$, $b=8$, and $c=9$: $\operatorname{cos} \alpha = \frac{8^2 + 9^2 - 7^2}{2(8)(9)} = \frac{96}{144} \approx 0.667$. Thus, $\alpha = \operatorname{cos}^{-1}(0.667) \approx 48.2^\circ$.
• The Law of Cosines is unambiguous because $\operatorname{cos}^{-1}(x)$ provides a unique angle between $0^\circ$ and $180^\circ$, unlike the Law of Sines which can have two possible angles.

Explanation

Think of this as the Pythagorean theorem with an upgrade for any triangle, not just right triangles. It is essential for solving triangles when you know Side-Angle-Side (SAS) or Side-Side-Side (SSS), where the Law of Sines cannot be used.

Property

Given two sides and the included angle (SAS), find the measures of the remaining side and angles of a triangle.

1. Sketch the triangle and identify knowns and unknowns.
2. Apply the Law of Cosines to find the length of the unknown side.
3. Apply the Law of Sines or Cosines to find the measure of a second angle.
4. Compute the measure of the remaining angle by subtracting the known angles from $180^\circ$.

Examples

• Given $a=15$, $c=20$, and $\beta=45^\circ$. Find side $b$: $b^2 = 15^2+20^2 - 2(15)(20)\operatorname{cos}(45^\circ) = 625 - 300\sqrt{2}$. So, $b \approx 14.19$.
• Continuing from above, find angle $\alpha$ using the Law of Sines: $\frac{\sin \alpha}{15} = \frac{\sin 45^\circ}{14.19}$. This gives $\sin \alpha \approx 0.747$, so $\alpha \approx 48.3^\circ$.
• Finally, find the third angle: $\gamma = 180^\circ - 45^\circ - 48.3^\circ = 86.7^\circ$. The triangle is now fully solved.

Explanation

When given a Side-Angle-Side setup, your first step is always the Law of Cosines to find the third side. Once you have that, the triangle can be fully solved using the Law of Sines to find the remaining angles.

Property

Given all three sides of a triangle (SSS), find the measures of the angles.

1. Use the rearranged Law of Cosines to find one of the angles. It is often best to find the largest angle first (opposite the longest side).
2. Use the Law of Sines or Cosines to find a second angle.
3. Find the third angle by subtracting the other two from $180^\circ$.

Examples

• Given $a=10$, $b=14$, and $c=18$. Find the largest angle, $\gamma$, opposite side $c$. $\operatorname{cos} \gamma = \frac{10^2+14^2-18^2}{2(10)(14)} = \frac{-28}{280} = -0.1$. So $\gamma \approx 95.7^\circ$.
• Continuing from above, find a second angle, say $\alpha$: $\frac{\sin \alpha}{10} = \frac{\sin 95.7^\circ}{18}$. This gives $\sin \alpha \approx 0.553$, so $\alpha \approx 33.6^\circ$.
• The final angle is $\beta = 180^\circ - 95.7^\circ - 33.6^\circ = 50.7^\circ$. The triangle is solved.

Explanation

If you have all three sides but no angles, the Law of Cosines is your tool to unlock the triangle. By solving for the cosine of an angle, you can find one angle and then easily determine the rest.

Property

Heron’s formula finds the area of oblique triangles in which sides $a$, $b$, and $c$ are known.

where $s = \frac{a+b+c}{2}$ is one half of the perimeter of the triangle, sometimes called the semi-perimeter.

Examples

• Find the area of a triangle with sides $a=13$, $b=14$, and $c=15$. The semi-perimeter is $s = \frac{13+14+15}{2} = 21$. The area is $\sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21(8)(7)(6)} = \sqrt{7056} = 84$ square units.
• A triangular garden has sides of 20 ft, 25 ft, and 30 ft. First, $s = \frac{20+25+30}{2} = 37.5$. The area is $\sqrt{37.5(17.5)(12.5)(7.5)} \approx 248$ square feet.
• Calculate the area of a triangular plot with sides 50 m, 70 m, and 80 m. The semi-perimeter is $s = \frac{50+70+80}{2} = 100$. The area is $\sqrt{100(50)(30)(20)} \approx 1732.1$ square meters.

Explanation

This is a powerful formula for finding a triangle's area using only its side lengths. You don't need to know any angles or the height, making it perfect for problems where only side measurements are available.