The area of a triangle refers to the total space or region enclosed by its three sides. Trigonometry provides a powerful method to calculate this area, especially when the height of the triangle is not available, and instead, we have side lengths and angles.
The basic formula for the area of a triangle is:
However, when the height is unknown, trigonometric methods can be used to find the area.
For a triangle with sides a, b and c given as below:
If a and b are two sides of a triangle and C is the included angle between them:
Area =
\frac{1}{2}\mathbf{ab} \sin(C)
Similarly,
- Area =
\frac{1}{2}\mathbf{bc} \sin(A) - Area =
\frac{1}{2}\mathbf{ca} \sin(B)
Derivation
Consider a triangle with sides a and b, and the angle between them is C. Drop a perpendicular from the vertex opposite to the side C, dividing the triangle into two right triangles.
Using
Area(ABC) = (1/2) × a × h . . . (i)
In triangle ADC, using the sine of angle C,
sin C = h/b
⇒
Now, substitute h =
⇒
Thus, the area is:
\text{Area} = \frac{1}{2} \times a \times b \times \sin(C)
This formula applies when two sides of the triangle and the angle between them are known.
Solved Examples
Example 1: Find the area of a triangle where a = 7 cm, b = 9 cm, and the angle between them Sin C =45°
- Given: a =7cm , b=9cm and sin c =45°
- Using the formula:
\text{Area} = \frac{1}{2} \times 7 \times 9 \times \sin(45^\circ) = \frac{1}{2} \times 7 \times 9 \times \frac{\sqrt{2}}{2}
\text{Area} = 31.5 \times \frac{\sqrt{2}}{2} \approx 22.27 \, \text{cm}^2 Thus, the area of the triangle is approximately
22.27 \, \text{cm}^2 .
Example 2: Find the area of a triangle where
- Given: a = 12
\text{cm} ,b = 8 \, \text{cm} , and\sin(60^\circ) = \frac{\sqrt{3}}{2} . - Using the formula:
\text{Area} = \frac{1}{2} \times 12 \times 8 \times \sin(60^\circ) = \frac{1}{2} \times 12 \times 8 \times \frac{\sqrt{3}}{2}
\text{Area} = 48 \times \frac{\sqrt{3}}{2} \approx 41.57 \, \text{cm}^2 Thus, the area of the triangle is approximately
41.57 \, \text{cm}^2 .
Example 3: Calculate the area of a triangle with sides a
- Given:
a = 14 \, \text{cm} ,b = 10 \, \text{cm} , and\sin(90^\circ) = 1 . - Using the formula:
\text{Area} = \frac{1}{2} \times 14 \times 10 \times 1 = 70 \, \text{cm}^2 Thus, the area of the triangle is
70 \, \text{cm}^2 .
Example 4: Find the area of a triangle where
- Given:
a = 16 \, \text{cm} , b = 12 \, \text{cm} , and \\sin(30^\circ) = \frac{1}{2} . - Using the formula:
\text{Area} = \frac{1}{2} \times 16 \times 12 \times \sin(30^\circ) = \frac{1}{2} \times 16 \times 12 \times \frac{1}{2}
\text{Area} = 48 \, \text{cm}^2 Thus, the area of the triangle is
48 \, \text{cm}^2 .
Example 5: A triangle has sides a = 9
- Given:
a = 9 \, \text{cm} ,b = 15 \, \text{cm} , and\sin(120^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} . - Using the formula:
\text{Area} = \frac{1}{2} \times 9 \times 15 \times \sin(120^\circ) = \frac{1}{2} \times 9 \times 15 \times \frac{\sqrt{3}}{2}
\text{Area} = 67.5 \times \frac{\sqrt{3}}{2} \approx 58.44 \, \text{cm}^2 Thus, the area of the triangle is approximately
58.44 \, \text{cm}^2 .
Practice Problems
Q 1. Find the area of a triangle with a
Q 2. Calculate the area of a triangle where a
Q 3. A triangle has sides a
Q 4. Determine the area of a triangle with a
Q 5. Find the area of a triangle with sides a
