Finding the Area of a Triangle with a Side and Two Angles

In this tutorial, we will learn how to find the area of a triangle when the length of one side and the two angles adjacent to it are known. This method uses the Law of Sines and basic trigonometry to compute the area.


Formula

The area of a triangle with a known side \(a\) and adjacent angles \(B\) and \(C\) is given by:

\[ \text{Area} = \frac{a^2 \sin(B) \sin(C)}{2 \sin(A)} \]

Explanation of Terms:

  • \(a\): The length of the known side.
  • \(B, C\): The two angles adjacent to the side \(a\).
  • \(A = 180^\circ – B – C\): The third angle of the triangle.
  • \(\sin\): The sine function, used to calculate the trigonometric ratio for an angle.


Examples

Example 1: Find the area of a triangle with side \(a = 8\) and angles \(B = 45^\circ\), \(C = 60^\circ\).

Given: \(a = 8\), \(B = 45^\circ\), \(C = 60^\circ\)

First, calculate the third angle:

\[ A = 180^\circ – B – C = 180^\circ – 45^\circ – 60^\circ = 75^\circ \]

Next, substitute the values into the formula:

\[ \text{Area} = \frac{a^2 \sin(B) \sin(C)}{2 \sin(A)} \] \[ \text{Area} = \frac{8^2 \cdot \sin(45^\circ) \cdot \sin(60^\circ)}{2 \cdot \sin(75^\circ)} \]

Calculate the sines:

\[ \sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707 \]

\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866\]

\[ \sin(75^\circ) \approx 0.966 \]

Substitute the sine values:

\[ \text{Area} = \frac{8^2 \times 0.707 \times 0.866}{2 \times 0.966} \]

Simplify:

\[ \text{Area} = \frac{64 \times 0.707 \times 0.866}{1.932} = \frac{39.18}{1.932} \approx 20.28 \]

Answer: The area of the triangle is approximately 20.28 square units.


Example 2: Find the area of a triangle with side \(a = 10\) and angles \(B = 50^\circ\), \(C = 70^\circ\).

Given: \(a = 10\), \(B = 50^\circ\), \(C = 70^\circ\)

First, calculate the third angle:

\[ A = 180^\circ – B – C = 180^\circ – 50^\circ – 70^\circ = 60^\circ \]

Next, substitute the values into the formula:

\[ \text{Area} = \frac{a^2 \sin(B) \sin(C)}{2 \sin(A)} \] \[ \text{Area} = \frac{10^2 \cdot \sin(50^\circ) \cdot \sin(70^\circ)}{2 \cdot \sin(60^\circ)} \]

Calculate the sines:

\[ \sin(50^\circ) \approx 0.766, \quad \sin(70^\circ) \approx 0.940, \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \]

Substitute the sine values:

\[ \text{Area} = \frac{10^2 \times 0.766 \times 0.940}{2 \times 0.866} \]

Simplify:

\[ \text{Area} = \frac{100 \times 0.766 \times 0.940}{1.732} = \frac{72}{1.732} \approx 41.57 \]

Answer: The area of the triangle is approximately 41.57 square units.