Applications of Triangle Solving

What You’ll Learn

In this lesson you’ll learn how to combine the Law of Sines and Law of Cosines to solve real-world problems involving non-right triangles.

The Concept

Now that you have both laws in your toolkit, the strategy for solving any triangle comes down to what information you’re given:

Given	Use
AAS or ASA (two angles + one side)	Law of Sines
SSA (two sides + non-included angle)	Law of Sines (watch for ambiguous case)
SAS (two sides + included angle)	Law of Cosines
SSS (all three sides)	Law of Cosines

The general approach for application problems:

1. Draw and label a triangle from the problem description
2. Identify what you know and what you need
3. Pick the right law based on the table above
4. Solve, then check that your answer makes sense

Sometimes you’ll need both laws in the same problem: use the Law of Cosines to find a side, then the Law of Sines to find a remaining angle.

Worked Example

Example 1: Surveying (AAS)

Two surveyors stand 500 meters apart at points A and B. They both sight a landmark at point C. Surveyor A measures the angle at A as 42°, and surveyor B measures the angle at B as 73°. How far is the landmark from each surveyor?

First, find angle C:

$$C = 180° - 42° - 73° = 65°$$

This is AAS, so use the Law of Sines. The side opposite C is AB = 500 m.

Find side a (distance from B to C):

$$\frac{a}{\sin A} = \frac{500}{\sin C}$$ $$a = \frac{500 \times \sin 42°}{\sin 65°} = \frac{500 \times 0.6691}{0.9063} \approx 369.1 \text{ m}$$

Find side b (distance from A to C):

$$b = \frac{500 \times \sin 73°}{\sin 65°} = \frac{500 \times 0.9563}{0.9063} \approx 527.6 \text{ m}$$

Example 2: Navigation (SAS)

A ship sails 15 km due north, then turns 35° to the east and sails another 10 km. How far is the ship from its starting point?

The two legs form two sides of a triangle (15 km and 10 km) with an included angle. The turn of 35° east of north means the interior angle of the triangle is 180° - 35° = 145°.

Using the Law of Cosines:

$$d^2 = 15^2 + 10^2 - 2(15)(10)\cos(145°)$$ $$d^2 = 225 + 100 - 300(-0.8192) = 325 + 245.8 = 570.8$$ $$d \approx \sqrt{570.8} \approx 23.9 \text{ km}$$

Example 3: Finding area with SAS

You can also find the area of any triangle when you know two sides and the included angle:

$$\text{Area} = \frac{1}{2}ab\sin C$$

A triangular plot of land has two sides measuring 120 m and 85 m with an included angle of 62°. Find the area.

$$\text{Area} = \frac{1}{2}(120)(85)\sin(62°) = \frac{1}{2}(10200)(0.8829) \approx 4503 \text{ m}^2$$

That’s about 1.11 acres.

Real-World Application

Triangle solving with these two laws shows up everywhere:

• Surveying and cartography (measuring distances to inaccessible points)
• Navigation (calculating position after course changes)
• Real estate (finding the area of irregular plots)
• Search and rescue (triangulating a signal from two known positions)
• Astronomy (measuring distances to stars using parallax)
• Engineering (analyzing forces in non-rectangular structures)

Example: search and rescue teams receive a distress signal. Two stations 20 miles apart each measure the angle to the signal. Using the Law of Sines, they triangulate the exact position.

You’ve Got This

Between the Law of Sines and the Law of Cosines, you can solve any triangle. The key is matching the right law to the information you have. Draw the triangle, label everything, pick your law, and solve. These two formulas are the complete toolkit for oblique triangles.

Quiz

Two sides of a triangle are 12 and 9 with an included angle of 50°. Which law do you use to find the third side?

• A.Law of Sines
• B.Law of Cosines
• C.Pythagorean Theorem
• D.SOH CAH TOA

Two surveyors 200 m apart measure angles of 55° and 80° to a tower. The angle at the tower is:

• A.$100°$
• B.$55°$
• C.$80°$
• D.$45°$

The area formula $\frac{1}{2}ab\sin C$ requires:

• A.Two sides and the included angle
• B.All three sides
• C.Two angles and one side
• D.Only the base and height

A ship sails 20 km north then turns 40° east and sails 15 km. The interior triangle angle is:

• A.$180°$
• B.$40°$
• C.$140°$
• D.$90°$

A triangle has sides 6, 8, and 11. To find angle A (opposite side 6), you'd compute:

• A.$\cos A = \frac{6^2 + 8^2 - 11^2}{2(6)(8)}$
• B.$\cos A = \frac{8^2 + 11^2 - 6^2}{2(8)(11)}$
• C.$\sin A = \frac{6}{11}$
• D.$\tan A = \frac{6}{8}$