What if I know all three sides but not any angles?

Use Heron's formula, which computes area from three sides alone without needing any angle. You can also find one angle with the Law of Cosines, then apply this formula, but Heron's method is more direct.

Why is there a sine in the formula instead of cosine?

Sine gives the perpendicular component of one side relative to the other—that is, the height of the triangle. Cosine gives the parallel component, which does not contribute to area, so using it would give the wrong result.

Can I use this formula if I know one side and two angles?

Not directly; this formula needs two sides and their included angle. If you have one side and two angles, first use the Law of Sines to find a second side, then apply this formula.

What is the relationship between this formula and the base-height formula?

This formula is equivalent to A = \frac{1}{2} \times \text{base} \times \text{height} when you recognize that b\sin C is the perpendicular height from angle C to side a. It saves you the step of finding height geometrically.

Area of a Triangle with Sine

Find the area of a triangle when you know two sides and the included angle—much faster than using just base and height.

A = \frac{1}{2}ab\sin C

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What each symbol means

What Area of a Triangle with Sine takes

Symbol	Meaning
$A$	The area of the triangle in square units (cm², m², etc.); confusing it with angle $A$ creates a mismatch between linear and angular quantities.
$a$	One side's length in linear units (meters, inches, etc.); using a non-adjacent side or reading it as angle $a$ invalidates the formula.
$b$	The other side's length in the same units as $a$ ; mistaking it for the height or using the wrong side breaks the calculation.
$C$	The included angle between $a$ and $b$ , measured in degrees or radians; using angle $A$ or angle $B$ gives the wrong sine value.

When to use it

Use this formula when you know the lengths of two sides of a triangle and the measure of the angle between them.

Level

Usually taught in: Pre-Calculus

Worked examples

1. Find area with two sides and an included angle

Problem

A triangle has sides of length 5 cm and 8 cm with an included angle of 60°. Find the area.

$A = \frac{1}{2} \cdot 5 \cdot 8 \cdot \sin(60°)$
Substitute $a = 5$ cm, $b = 8$ cm, and $C = 60°$ into the formula.
$A = \frac{1}{2} \cdot 5 \cdot 8 \cdot \frac{\sqrt{3}}{2}$
Evaluate $\sin(60°) = \frac{\sqrt{3}}{2}$ .
$A = \frac{40\sqrt{3}}{4}$
Multiply: $\frac{1}{2} \cdot 5 \cdot 8 = 20$ , then $20 \cdot \frac{\sqrt{3}}{2} = \frac{40\sqrt{3}}{4}$ .
$A = 10\sqrt{3} \text{ cm}^2$
Simplify: $\frac{40\sqrt{3}}{4} = 10\sqrt{3}$ square cm.

Answer: $A = 10\sqrt{3} \text{ cm}^2 \approx 17.3 \text{ cm}^2$

We substituted directly into the formula using the two known sides and their included angle. Since $\sin(60°)$ is a standard value, we simplified to an exact answer.

2. Find area with radian measure and decimal sides

Problem

A triangle has sides of 3.5 and 4.2 units with an included angle of

\frac{\pi}{4}

radians. Find the area to two decimal places.

$A = \frac{1}{2} \cdot 3.5 \cdot 4.2 \cdot \sin\left(\frac{\pi}{4}\right)$
Substitute $a = 3.5$ units, $b = 4.2$ units, and $C = \frac{\pi}{4}$ into the formula.
$A = \frac{1}{2} \cdot 3.5 \cdot 4.2 \cdot \frac{\sqrt{2}}{2}$
Evaluate $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ .
$A = \frac{1}{2} \cdot 14.7 \cdot \frac{\sqrt{2}}{2}$
Multiply: $3.5 \times 4.2 = 14.7$ .
$A = \frac{14.7\sqrt{2}}{4}$
Combine the fractions: $\frac{1}{2} \cdot 14.7 \cdot \frac{\sqrt{2}}{2} = \frac{14.7\sqrt{2}}{4}$ .
$A \approx \frac{14.7 \times 1.414}{4} \approx 5.20 \text{ square units}$
Use $\sqrt{2} \approx 1.414$ and calculate: $\frac{20.79}{4} \approx 5.20$ .

Answer: $A \approx 5.20 \text{ square units}$

With decimal side lengths and a radian angle, exact arithmetic became complicated. Using decimal approximations for $\sqrt{2}$ let us arrive at a practical answer.

3. Find area of a basketball training zone

Problem

During basketball practice, a coach sets up a training zone on a triangular part of the court using two ropes of 12 meters and 15 meters, with an angle of 45° between them. What is the area of this training zone?

$A = \frac{1}{2} \cdot 12 \cdot 15 \cdot \sin(45°)$
Identify the given values: $a = 12$ m, $b = 15$ m, and $C = 45°$ .
$A = \frac{1}{2} \cdot 12 \cdot 15 \cdot \frac{\sqrt{2}}{2}$
Evaluate $\sin(45°) = \frac{\sqrt{2}}{2}$ .
$A = \frac{1}{2} \cdot 180 \cdot \frac{\sqrt{2}}{2}$
Multiply: $12 \times 15 = 180$ .
$A = \frac{180\sqrt{2}}{4}$
Combine the fractions: $\frac{1}{2} \cdot 180 \cdot \frac{\sqrt{2}}{2} = \frac{180\sqrt{2}}{4}$ .
$A = 45\sqrt{2} \text{ m}^2$
Simplify: $\frac{180\sqrt{2}}{4} = 45\sqrt{2}$ square meters.
$A \approx 45 \times 1.414 \approx 63.63 \text{ m}^2$
Use $\sqrt{2} \approx 1.414$ to find the decimal approximation.

Answer: $A = 45\sqrt{2} \text{ m}^2 \approx 63.63 \text{ m}^2$

In real-world scenarios like sports court setup, measuring two sides and the angle between them is often easier than finding perpendicular height. This formula gives both an exact symbolic answer and a practical decimal result.

Common mistakes

Where Area of a Triangle with Sine usually goes wrong

The mistake	Why it is wrong	The fix
Writing $A = \frac{1}{2}ab\sin(A)$ instead of $A = \frac{1}{2}ab\sin(C)$	Angle $A$ is opposite side $a$ , not between sides $a$ and $b$ ; using the wrong angle gives a completely different sine value.	Identify which angle sits at the vertex between your two known sides—that's always the one you must use.
Writing $A = ab\sin C$ and omitting the $\frac{1}{2}$ factor	The factor of $\frac{1}{2}$ is essential because area equals half of base times height, and $b\sin C$ represents the height; dropping it doubles your answer.	Always include $\frac{1}{2}$ in the formula, or recall that area is never just base times height in a triangle.
Entering the angle in degrees when the calculator is in radian mode, or vice versa	If your calculator is set to radian mode but you input a degree angle without converting, sine returns a nonsense value; the modes must match.	Check your calculator's mode before computing sine, or convert explicitly: $60° = \frac{\pi}{3}$ radians and $45° = \frac{\pi}{4}$ radians.

Tips and when to use something else

The signal to use this formula is 'two sides and the included angle'—if you have that setup, reach for this before other methods.
If you know all three sides but no angles, use Heron's formula instead, which requires no sine at all.
Radian and degree mode on your calculator matter critically; a mode mismatch produces sine values that are completely wrong.
This formula is the practical version of the vector cross-product formula $A = \frac{1}{2}|ab\sin\theta|$ for area.

Related formulas

Law of CosinesFinds unknown sides or angles in any triangle when you know two sides and the angle between them, or all three sides.Law of SinesLaw of Sines connects side lengths to their opposite angles in a triangle, used to find unknown sides or angles when an angle-side pair is known.Sine Cosine TangentFind missing sides or angles in right triangles using sine, cosine, and tangent ratios—essential for SAT and ACT geometry.

Frequently asked questions

What if I know all three sides but not any angles?: Use Heron's formula, which computes area from three sides alone without needing any angle. You can also find one angle with the Law of Cosines, then apply this formula, but Heron's method is more direct.
Why is there a sine in the formula instead of cosine?: Sine gives the perpendicular component of one side relative to the other—that is, the height of the triangle. Cosine gives the parallel component, which does not contribute to area, so using it would give the wrong result.
Can I use this formula if I know one side and two angles?: Not directly; this formula needs two sides and their included angle. If you have one side and two angles, first use the Law of Sines to find a second side, then apply this formula.
What is the relationship between this formula and the base-height formula?: This formula is equivalent to $A = \frac{1}{2} \times \text{base} \times \text{height}$ when you recognize that $b\sin C$ is the perpendicular height from angle $C$ to side $a$ . It saves you the step of finding height geometrically.

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Reviewed 2026-09-18