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Deriving the
Triangle Area Formula

Discover how trigonometry unlocks a powerful formula: \(\frac{1}{2}ab\sin C\). Move beyond base-times-height and find the area of any triangle using just two sides and the included angle.

Precalculus

Why Do We Need a New Formula?

You already know that the area of a triangle is \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\). But what happens when you don't know the height?

side \(b\)
side \(a\)
\(h=?\)
\(C\)

Recall: The Sine Ratio

In a right triangle, sine relates the opposite side to the hypotenuse:

adjacent
opposite
hypotenuse
\(\theta\)

This relationship is the key to expressing height in terms of an angle, which is the heart of our derivation.

Setting Up the Derivation

Consider triangle \(ABC\) with sides \(a\), \(b\), and included angle \(C\). Drop a perpendicular from vertex \(B\) to side \(AC\), creating height \(h\).

\(A\)
\(B\)
\(C\)
\(a\)
\(c\)
\(b\)
\(h\)
\(A\)
\(C\)

Completing the Derivation

1
Start with the classic formula: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} b h\)
2
Substitute \(h = a\sin C\): \(\text{Area} = \frac{1}{2} b (a\sin C) = \frac{1}{2}ab\sin C\)
3
Result: The area depends only on two sides and the included angle: no height needed!

Three Equivalent Forms

Since you can drop the height from any vertex, the formula works with any pair of sides and their included angle:

\(\text{Area} = \frac{1}{2}\,ab\sin C\)
Sides \(a\), \(b\) with included angle \(C\)
\(\text{Area} = \frac{1}{2}\,bc\sin A\)
Sides \(b\), \(c\) with included angle \(A\)
\(\text{Area} = \frac{1}{2}\,ac\sin B\)
Sides \(a\), \(c\) with included angle \(B\)

Classic vs. Trig Area Formula

\(\frac{1}{2}bh\) \(\frac{1}{2}ab\sin C\)
Requirements Base and perpendicular height Two sides and included angle
Height needed? Yes No
Works for oblique triangles? Only if you can find \(h\) Always
Best when Height is given or easy to find SAS data is given

Why Does \(\sin C\) Matter?

The sine of the included angle controls how much "opening" there is between the two sides. As the angle changes, the area changes:

\(30^\circ\)
 Small area
\(90^\circ\)
 Maximum area
\(150^\circ\)
 Small again

How to Apply the Formula

1
Identify the SAS data: locate two known sides and the angle between them.
2
Label clearly: call the sides \(a\) and \(b\), and the included angle \(C\).
3
Substitute into \(\text{Area} = \frac{1}{2}ab\sin C\).
4
Compute the sine value, multiply, and include square units in your answer.

Example 1: Basic SAS

Find the area of a triangle with sides \(a = 8\) cm, \(b = 5\) cm, and included angle \(C = 60^\circ\).

\(a = 8\text{ cm}\)
\(b = 5\text{ cm}\)
\(c\)
\(60^\circ\)
\(C\)
\(A\)
\(B\)

Example 2: Obtuse Angle

Find the area of a triangle with sides \(a = 10\), \(b = 7\), and included angle \(C = 120^\circ\).

\(c\)
\(a=10\)
\(b=7\)
\(120^\circ\)
\(B\)
\(C\)
\(A\)

Example 3: Finding a Missing Side

A triangle has area \(24\) cm², one side \(a = 8\) cm, and included angle \(C = 30^\circ\). Find the other side \(b\).

1
Write the formula: \(24 = \frac{1}{2}(8)(b)\sin 30^\circ\)
2
Simplify: \(24 = \frac{1}{2}(8)(b)\left(\frac{1}{2}\right) = 2b\)
3
Solve: \(b = \frac{24}{2} = 12\) cm

Example 4: Real-World Application

A surveyor measures two sides of a triangular plot as 150 m and 200 m, with an included angle of \(75^\circ\). Find the area of the plot.

\(150\text{ m}\)
\(200\text{ m}\)
\(75^\circ\)

Common Mistakes

✕ Wrong Angle
Using an angle that is not between the two chosen sides. The angle must be the included angle.
✕ Forgetting \(\frac{1}{2}\)
Writing \(ab\sin C\) instead of \(\frac{1}{2}ab\sin C\). This doubles the correct area.
✕ Degrees vs. Radians
Calculating \(\sin C\) with the calculator set to radians when the angle is given in degrees (or vice versa).

Formula Summary

Scenario Formula When to Use
Height known \(\frac{1}{2}bh\) Base and height given
SAS (two sides + included angle) \(\frac{1}{2}ab\sin C\) Most common exam scenario
All three sides (SSS) Heron's formula No angles given at all
Right triangle \(\frac{1}{2}\,\text{leg}_1 \times \text{leg}_2\) Special case of \(\frac{1}{2}ab\sin 90^\circ\)

Question 1 of 5

What is the area of a triangle with sides \(a = 6\), \(b = 10\), and included angle \(C = 90^\circ\)?

Question 2 of 5

A triangle has sides \(a = 12\) and \(b = 9\) with included angle \(C = 30^\circ\). What is its area?

Question 3 of 5

In the formula \(\text{Area} = \frac{1}{2}ab\sin C\), what does \(C\) represent?

Question 4 of 5

A triangle has area \(18\), one side \(a = 6\), and included angle \(C = 45^\circ\). Find the other side \(b\).

Question 5 of 5

For two fixed side lengths, at what included angle \(C\) is the triangle's area maximized?