10.1 Non-right Triangles: Law of Sines

Law of Sines and Non-right Triangles

Law of Sines for Non-right Triangles

The Law of Sines lets you solve triangles that don't contain a right angle. It works by setting up a proportion: in any triangle, the ratio of a side length to the sine of its opposite angle is the same for all three side-angle pairs.

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

Here, $A$, $B$, and $C$ are the three angles, and $a$, $b$, $c$ are the sides opposite those angles respectively. You can also write this flipped (sides on top), which is sometimes easier when solving for a side:

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

To use the Law of Sines, you need at least one complete ratio (a known side and its opposite angle). From there, you can solve for unknowns depending on what information you're given.

AAS or ASA (two angles and a side):

1. Use the fact that angles in a triangle sum to 180° to find the third angle.
2. Set up a Law of Sines proportion using the known side and its opposite angle.
3. Solve for each unknown side.

For example, if you know $A = 40°$, $B = 60°$, and $a = 10$, first find $C = 180° - 40° - 60° = 80°$. Then solve $\frac{10}{\sin 40°} = \frac{b}{\sin 60°}$ to get $b$.

SSA (two sides and an angle opposite one of them):

1. Use the Law of Sines to find the sine of the unknown angle.
2. Check for the ambiguous case (see below).
3. Once angles are determined, find the remaining side with another proportion.

The Ambiguous Case (SSA) is the trickiest part of the Law of Sines. When you're given two sides and a non-included angle, there may be:

• No triangle if the sine value you calculate is greater than 1 (impossible for sine).
• Exactly one triangle if the sine value gives an angle where only one configuration works (e.g., the angle must be acute, or the known angle is already obtuse).
• Two triangles if the sine value gives an acute angle $\theta$, and the supplement $180° - \theta$ also produces a valid triangle (meaning the angles still sum to less than 180°).

Always check both possible angles when working an SSA problem. Plug each into the angle sum to see if it's valid.

Area Calculation with the Sine Function

When you know two sides and the angle between them (SAS), you can find the area without needing a height measurement:

$\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin C$

Here $a$ and $b$ are two side lengths and $C$ is the included angle (the angle formed between those two sides). This formula works because $b \cdot \sin C$ effectively calculates the height of the triangle relative to side $a$ as the base.

If you aren't directly given SAS information, you can often get there:

1. If given AAS, find the third angle (angles sum to 180°), then use the Law of Sines to find a missing side so you have two sides and their included angle.
2. Then apply the area formula above.

For example, if a triangular plot of land has two sides measuring 150 ft and 200 ft with an included angle of 35°, the area is $\frac{1}{2}(150)(200)\sin 35° \approx 8{,}604 \text{ ft}^2$.

Real-World Applications of the Law of Sines

Many practical problems involve triangles that aren't right triangles. The general approach:

1. Sketch the triangle and label all known sides and angles from the problem.
2. Identify what combination you have: AAS, ASA, SSA, or SAS.
3. Apply the Law of Sines to find missing sides or angles. If you have SAS and need a missing angle first, you'll actually need the Law of Cosines (covered in 10.2), or you can jump straight to the area formula if that's what the problem asks for.
4. Interpret your answer in context. Side lengths must be positive, and angles in a triangle must be between 0° and 180°.

A common application: you're standing a known distance from a tall object, you measure the angle of elevation, and you have one more piece of information (like a second angle from a different point). Setting up the triangle and applying the Law of Sines gives you the object's height.

Deriving the Law of Sines

The Law of Sines can be derived by dropping an altitude from any vertex to the opposite side. If you drop a height $h$ from vertex $C$ to side $c$, then $h = a \sin B$ and $h = b \sin A$. Setting these equal gives $a \sin B = b \sin A$, which rearranges to $\frac{\sin A}{a} = \frac{\sin B}{b}$. Repeating with a different altitude completes the full proportion.

There's also an elegant connection to the circumscribed circle: each ratio $\frac{a}{\sin A}$ equals $2R$, where $R$ is the radius of the circle that passes through all three vertices of the triangle.