8.4 Law of Sines and Law of Cosines

Solving Oblique Triangles

Oblique triangles are triangles that don't contain a right angle, which means SOH-CAH-TOA and the Pythagorean Theorem won't work directly. The Law of Sines and Law of Cosines let you solve any triangle by relating its sides and angles. Knowing which law to reach for (and why) is the key skill in this section.

Law of Sines

The Law of Sines connects each side of a triangle to the sine of its opposite angle:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Here, $a$, $b$, and $c$ are the side lengths, and $A$, $B$, $C$ are the angles opposite those sides. This works for any triangle, not just right triangles.

Solving for a missing side:

1. Identify a complete pair you already know (a side and its opposite angle).
2. Set up a proportion with the missing side and its opposite angle.
3. Cross-multiply and solve.

Example: In triangle ABC, you know $A = 40°$, $B = 60°$, and $a = 10$. Find $b$.

$\frac{10}{\sin 40°} = \frac{b}{\sin 60°}$

$b = \frac{10 \cdot \sin 60°}{\sin 40°} \approx \frac{10 \cdot 0.866}{0.643} \approx 13.47$

Solving for a missing angle:

1. Identify a complete pair (side + opposite angle).
2. Set up a proportion with the known side whose opposite angle is missing.
3. Solve for the sine of the unknown angle, then use inverse sine ($\sin^{-1}$) to find the angle.

Watch out: the inverse sine function only gives you an acute angle. If the missing angle could be obtuse, you need to check whether the supplement ($180° - \theta$) also produces a valid triangle (meaning the angles still add to $180°$). This is the ambiguous case (SSA), and it can produce zero, one, or two valid triangles.

Law of Cosines

The Law of Cosines is a generalization of the Pythagorean Theorem that works for any triangle:

$c^2 = a^2 + b^2 - 2ab \cos C$

The variable $C$ is the angle between sides $a$ and $b$, and $c$ is the side opposite that angle. Notice that if $C = 90°$, then $\cos 90° = 0$ and the formula reduces to $c^2 = a^2 + b^2$. That's how you know this formula is consistent with what you already learned about right triangles.

Solving for a missing side (SAS):

1. Identify the two known sides and the included angle between them.

2. Plug into the formula: $c^2 = a^2 + b^2 - 2ab\cos C$.

3. Compute the right side, then take the square root to get $c$.

Example: Given $a = 7$, $b = 9$, and $C = 120°$.

$c^2 = 7^2 + 9^2 - 2(7)(9)\cos 120°$

$c^2 = 49 + 81 - 126(-0.5) = 130 + 63 = 193$

$c = \sqrt{193} \approx 13.89$

Notice that $\cos 120°$ is negative, which turns the subtraction into addition. This makes $c$ longer than it would be in a right triangle with the same two sides. That makes geometric sense: an obtuse angle "pushes" the opposite side out farther.

Solving for a missing angle (SSS):

When you know all three sides, rearrange the formula to isolate the cosine of the unknown angle:

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Then use $\cos^{-1}$ to find the angle. Unlike inverse sine, inverse cosine gives a unique answer between $0°$ and $180°$, so there's no ambiguous case here. This is one reason many students prefer to switch to the Law of Cosines (once they have enough information) to find remaining angles after using the Law of Sines.

Choosing the Right Law

The information you're given determines which law to use:

A quick rule of thumb: if you have a complete side-angle pair to start with, use the Law of Sines. If you don't (because you have SAS or SSS), use the Law of Cosines.

Also remember: once you find one missing piece, your situation changes. For example, after using the Law of Cosines to find a side in an SAS problem, you now have a complete side-angle pair and can switch to the Law of Sines for the remaining unknowns.

Applications in Real-World Problems

Surveying, navigation, and architecture problems all come down to the same process:

1. Sketch and label the triangle from the problem's description. Mark every known side and angle.
2. Classify what you have (AAS, SSA, SAS, or SSS) using the table above.
3. Apply the appropriate law and solve for the unknown.
4. Interpret your answer in context. If the problem asks for a distance between two landmarks, make sure your answer corresponds to the correct side of the triangle.

Example scenario: Two ranger stations are 15 km apart. Station A measures the angle to a fire at $54°$, and Station B measures it at $68°$. The third angle is $180° - 54° - 68° = 58°$. You now have ASA, so use the Law of Sines to find the distance from either station to the fire. The 15 km side is opposite the $58°$ angle (the angle at the fire), so your proportion would look like:

$\frac{15}{\sin 58°} = \frac{d}{\sin 68°}$

where $d$ is the distance from Station A to the fire. Solving gives $d \approx 16.39$ km.