How to Find the Area of a Triangle Using Trigonometry

Step-by-step Guide to How to Find the Area of a Triangle Using Trigonometry

Here is a step-by-step guide to how to find the area of a triangle using trigonometry:

Step 1: Understand the Basic Area Formula

For any triangle, the general area formula is:

\(Area \ =\frac{1}{2}​× \ base \ × \ height\)

However, if you don’t have a height, but have the lengths of two sides and the included angle between them, trigonometry comes to the rescue.

Step 2: Understand the Trigonometric Formula

Given a triangle with sides \(a\), \(b\), and \(c\), and the angle \(C\) opposite side \(c\), the area \(K\) can be found by: \(K=\frac{1}{2}​×a×b×sin(C)\)

Step 3: Determine the Relevant Sides and Angle

Identify two sides of the triangle and the angle between them (the included angle). Label these sides \(a\) and \(b\) and the angle as \(C\).

Step 4: Use the Sine Function

Once you’ve identified the two sides and the included angle, use the sine function to find the value of \(sin(C)\).

Step 5: Plug into the Formula

Insert the known values into the formula: \(K=\frac{1}{2}​×a×b×sin(C)\) Compute the result to get the area of the triangle.

Step 6: Units

Ensure you square the units you use for the side lengths since you’re finding an area. For instance, if side lengths are in centimeters, the area will be in square centimeters.

Examples:

Example 1:

Consider a triangle with sides \(a=8 \ cm\) and \(b=10 \ cm\), and an angle \(C=45°\). Evaluate the area of the triangle.

Solution:

1. Calculate the sine of the included angle: \(sin \ (45°)=\frac{\sqrt{2}}{2}\)
2. Using the formula: \(K=\frac{1}{2}​×\ a×b×sin(C)\)\(=\frac{1}{2}×8 \ cm×10 \ cm×\frac{\sqrt{2}}{2}\)

\(K=20 \sqrt{2} \ cm^2\)

Thus, the area of the triangle in this instance is \(K=20 \sqrt{2} \ cm^2\).

Example 2:

Consider a triangle with sides \(a=6 \ cm\) and \(b=9 \ cm\), and an angle \(C=30°\). Evaluate the area of the triangle.

Solution:

1. Calculate the sine of the included angle: \(sin \ (30°)=0.5\)
2. Using the formula: \(K=\frac{1}{2}​×\ a×b×sin(C)\)\(=\frac{1}{2}​×6  \ m×9 \ m×0.5\)

\(K=13.5 \ m^2\)

Thus, the area of the triangle in this scenario is \(13.5 \ m^2\).