Area of a Triangle Using Sine

15 min read

The grade-school area rule is $A = \frac{1}{2} \times \text{base} \times \text{height}$, and it works whenever the height is handed to you. Often it is not. A surveyor stands at one corner of a triangular plot, measures the two sides that meet there with a tape, and reads the angle between them off an instrument. There is no perpendicular height to measure, and dropping one out in a field is not practical. The same situation shows up for a triangular sail described by two edge lengths and the angle at the mast, or a triangular piece of sheet metal cut by two rulers and a protractor. In every one of these cases the known data is two sides and the angle between them, and the height is missing. Trigonometry supplies the missing height for free: the sine of the included angle turns one of the two known sides into the height the old formula wanted. The result is a single line, \[ A = \tfrac{1}{2}\,ab\sin C, \] where $a$ and $b$ are the two known sides and $C$ is the angle squeezed between them.

By the end of this page you can:

• Compute the area of a triangle from two sides and the included angle using $A = \frac{1}{2}ab\sin C$.
• Identify which angle is the included angle for a given pair of sides, and reject an angle that is not included.
• Apply the formula when the included angle is obtuse, where $\sin C$ is still positive.
• Solve the formula in reverse for a missing side or a missing angle when the area is known.
• Explain, from the altitude $h = a\sin C$, why the sine version is the same area rule as $\frac{1}{2}\,\text{base}\times\text{height}$.

The bare minimum: the area of a triangle is $A = \frac{1}{2}ab\sin C$, where $a$ and $b$ are two sides and $C$ is the angle between those two sides. The angle has to sit between the two sides you use, the one-half stays attached, and the sine handles the height. Hold those three facts and most area problems are one substitution away from done.

Quick Reference

Before You Start

Check each box you can do from memory. A box you cannot check yet points you to a quick refresher, not a grade.

• I can read the sine of an angle in a right triangle as the opposite side over the hypotenuse, and solve that relation for the opposite side. (Right Triangle Trigonometry)
• I can find $\sin 30^\circ$, $\sin 45^\circ$, $\sin 60^\circ$, and $\sin 90^\circ$ without a calculator.
• I can find the sine of an obtuse angle such as $150^\circ$ using its reference angle. (Reference Angles)

A 60-second self-check. Pick an answer, then reveal the reasoning.

Try This First

A triangle has two sides of length $6$ and $8$, and the angle caught between those two sides is exactly $90^\circ$. Find its area two ways: once with the grade-school $\frac{1}{2}\,\text{base}\times\text{height}$ rule, and once with $A = \frac{1}{2}ab\sin C$. Do they agree?

The Idea Behind the Sine Area Formula

Start with the rule everyone already knows: a triangle’s area is one half of its base times its height. The trouble is that the height is the perpendicular distance from one vertex straight down to the opposite side, and that perpendicular is rarely one of the measured sides. When the known data is two sides and the angle between them, the perpendicular height is missing.

Dropping an altitude recovers it. Label the triangle with sides $a$ and $b$ meeting at angle $C$, and use $b$ as the base. Drop a perpendicular from the far vertex to the base $b$. That perpendicular has some height $h$, and it forms a right triangle in which $a$ is the hypotenuse and $C$ is one of the acute angles. In that right triangle, \[ \sin C = \frac{h}{a}, \quad\text{so}\quad h = a\sin C. \]

Now substitute $h = a\sin C$ into $A = \frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}\,b\,h$: \[ A = \tfrac{1}{2}\,b\,(a\sin C) = \tfrac{1}{2}\,ab\sin C. \]

Nothing about the area rule changed. The sine simply computes the height that the data did not hand over. The angle $C$ has to be the angle between $a$ and $b$, because that is the angle that sits at the foot of the altitude in the right triangle. An angle somewhere else in the figure does not measure this height.

Prerequisite Hub

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Builds on:

Unlocks (what this section feeds into next):

Cross-course prerequisites: none.

Official Definitions

Area of a triangle from two sides and the included angle

The area $A$ of a triangle with sides of lengths $a$ and $b$ and included angle $C$ is \[ A = \tfrac{1}{2}\,ab\sin C. \]

Stewart 7e, Section 6.3 Trigonometric Functions of Angles (the area formula, with exercises on p. 499).

The phrase included angle names the angle formed where the two sides $a$ and $b$ meet. The two sides are the rays out of that vertex, and the included angle is the opening between them.

The altitude as side times sine

If an altitude of length $h$ is dropped to the side $b$ from the opposite vertex, and $C$ is the angle between sides $a$ and $b$, then $h = a\sin C$.

Derived in Stewart 7e from the right-triangle definition $\sin C = \dfrac{\text{opposite}}{\text{hypotenuse}}$ (Section 6.2).

The included angle for a pair of sides is the only angle that produces this height. For each pair of sides there is exactly one included angle, and the formula needs that one.

Worked Examples

Worked Example 1: Two sides and the included angle

Problem. A triangle has sides $a = 10$ and $b = 12$ with included angle $C = 30^\circ$. Find its area.

Predict first. If the angle were a full $90^\circ$, the area would be $\frac{1}{2}(10)(12)(1) = 60$. The angle here is only $30^\circ$, and $\sin 30^\circ = \frac{1}{2}$, so the area should come out to about half of $60$, near $30$. Hold that prediction.

Step 1. Write the formula and substitute. \[ A = \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(10)(12)\sin 30^\circ. \]

Step 2. Replace $\sin 30^\circ$ with its exact value $\frac{1}{2}$. \[ A = \tfrac{1}{2}(10)(12)\left(\tfrac{1}{2}\right). \]

Step 3. Multiply left to right. First $\frac{1}{2}(10)(12) = 60$, then $60 \times \frac{1}{2} = 30$. \[ A = 30. \]

Check the prediction. The area is $30$ square units, which is half of the $60$ from the $90^\circ$ case, exactly as predicted because $\sin 30^\circ = \frac{1}{2}$ is half of $\sin 90^\circ = 1$. The smaller the included angle, the flatter the triangle and the smaller the area.

Worked Example 2: An obtuse included angle

Problem. A triangular plot has two sides of length $a = 8$ meters and $b = 5$ meters meeting at an obtuse included angle $C = 150^\circ$. Find the area.

Predict first. The angle $150^\circ$ is obtuse, yet a triangle still has a real positive area, so $\sin 150^\circ$ must be positive. Its reference angle is $180^\circ - 150^\circ = 30^\circ$, so $\sin 150^\circ = \sin 30^\circ = \frac{1}{2}$. The area should match what a $30^\circ$ included angle would give on the same two sides. Hold that prediction.

Step 1. Substitute the sides and the angle. \[ A = \tfrac{1}{2}(8)(5)\sin 150^\circ. \]

Step 2. Use the reference angle. Since $150^\circ$ lies in the second quadrant where sine is positive, $\sin 150^\circ = \sin 30^\circ = \frac{1}{2}$. \[ A = \tfrac{1}{2}(8)(5)\left(\tfrac{1}{2}\right). \]

Step 3. Multiply. $\frac{1}{2}(8)(5) = 20$, then $20 \times \frac{1}{2} = 10$. \[ A = 10 \text{ square meters}. \]

Check the prediction. The area is positive, $10$ square meters, confirming that an obtuse included angle causes no trouble: $\sin C$ stays positive for every angle strictly between $0^\circ$ and $180^\circ$, which is the entire range of possible triangle angles. Replacing $150^\circ$ with its supplement $30^\circ$ gives the same sine and the same area.

Worked Example 3: An included angle that is not at a special value

Problem. A sail is a triangle with two edges of length $a = 4$ feet and $b = 7$ feet meeting at the mast, where the included angle is $C = 60^\circ$. Find the exact area, then a decimal estimate.

Predict first. With $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$, the area should be a bit less than the $90^\circ$ value $\frac{1}{2}(4)(7)(1) = 14$. Expect something near $12$. Hold that prediction.

Step 1. Substitute. \[ A = \tfrac{1}{2}(4)(7)\sin 60^\circ = \tfrac{1}{2}(4)(7)\cdot\frac{\sqrt{3}}{2}. \]

Step 2. Multiply the numbers, keeping $\sqrt{3}$ exact. First $\frac{1}{2}(4)(7) = 14$, then \[ A = 14 \cdot \frac{\sqrt{3}}{2} = 7\sqrt{3}. \]

Step 3. Estimate with $\sqrt{3} \approx 1.732$. \[ A = 7\sqrt{3} \approx 7(1.732) = 12.124 \text{ square feet}. \]

Check the prediction. The exact area is $7\sqrt{3}$ square feet, about $12.1$, slightly below the $14$ from a right angle, exactly as predicted because $\sin 60^\circ < 1$. Carrying $\sqrt{3}$ as an exact radical until the last step keeps the answer precise.

Worked Example 4: Solving in reverse for a missing side

Problem. A triangle has area $A = 45$ square units, one known side $b = 18$, and the included angle $C = 30^\circ$ between $b$ and the unknown side $a$. Find $a$.

Predict first. With $\sin 30^\circ = \frac{1}{2}$, the formula reads $A = \frac{1}{2}\,a\,(18)\left(\frac{1}{2}\right) = 4.5\,a$. So $a = A/4.5 = 45/4.5 = 10$. Hold that prediction and confirm it through the algebra.

Step 1. Write the formula with the known values in place. \[ 45 = \tfrac{1}{2}\,a\,(18)\sin 30^\circ. \]

Step 2. Replace $\sin 30^\circ = \frac{1}{2}$ and simplify the coefficient of $a$. \[ 45 = \tfrac{1}{2}\,a\,(18)\left(\tfrac{1}{2}\right) = \tfrac{18}{4}\,a = 4.5\,a. \]

Step 3. Divide both sides by $4.5$. \[ a = \frac{45}{4.5} = 10. \]

Check. Substitute $a = 10$ back in: $\frac{1}{2}(10)(18)\left(\frac{1}{2}\right) = \frac{1}{2}(10)(18) = 90$, then $90 \times \frac{1}{2} = 45$. The area returns to $45$, so $a = 10$ is correct, matching the prediction. The same formula runs forward to find area and backward to find a missing side or angle.

Common Misconceptions

Practice Problems

Mastery Checklist

Novice (Level 1-2):

• I can state the formula $A = \frac{1}{2}ab\sin C$ from memory and name what $a$, $b$, and $C$ stand for.
• I can compute the area when the included angle is $30^\circ$, $45^\circ$, $60^\circ$, or $90^\circ$ using exact sine values.
• I can keep the one-half attached and explain that it makes the triangle half of a parallelogram.

Competent (Level 3-4):

• I can give an exact radical answer such as $7\sqrt{3}$ and then a decimal estimate.
• I can apply the formula when the included angle is obtuse by using its reference angle.
• I can identify the included angle for a given pair of sides and reject an angle that is not included.

Proficient (Level 5):

• I can solve $A = \frac{1}{2}ab\sin C$ in reverse for a missing side or a missing angle.
• I can explain why a missing angle has two solutions, an angle and its supplement, that produce the same area.
• I can derive the formula from $h = a\sin C$ and the half-base-times-height rule.

Mental Model

Picture the two known sides as two rulers pinned together at one end, with the pin sitting at the included angle $C$. Spread them open and a triangle fills the gap between their far ends. The product $ab$ measures how large the two rulers are; the factor $\sin C$ measures how far open the angle is, running from $0$ when the rulers lie flat on each other up to $1$ when they stand at a right angle; and the one-half cuts the spanned parallelogram down to the triangle. Open the angle wider toward $90^\circ$ and the area grows to its maximum, then past $90^\circ$ the sine falls back and the area shrinks again, which is why $\sin 150^\circ$ and $\sin 30^\circ$ give the same area.

Connections

Looking back:

• Right Triangle Trigonometry supplies the relation $\sin C = \frac{\text{opposite}}{\text{hypotenuse}}$ that becomes the height $h = a\sin C$.
• Reference Angles gives the sine of an obtuse included angle through its acute reference angle.

Looking ahead:

• Law of Sines and Cosines extends triangle measurement to every missing side and angle, and the area formula is the first relation that ties two sides to the angle between them.

Real-world connections:

• Land surveying computes the area of a plot from two measured boundary lengths and the angle between them, with no need to pace off a perpendicular height.
• Sail and structural design uses the formula to find the area of a triangular sail or panel from two edge lengths and the angle at their shared corner.

Go Deeper (optional)

None of this is needed to compute an area or to pass the section. It is here for readers who want to see where the formula leads.

The formula is one half of a cross product. In vector terms, if the two sides $a$ and $b$ are vectors sharing a starting point, the magnitude of their cross product is $|a||b|\sin C$, which is exactly the area of the parallelogram they span. The triangle is half of that parallelogram, so $A = \frac{1}{2}|a||b|\sin C$ is the same statement written for vectors. This is the bridge from the trigonometry here to the area and volume computations of multivariable calculus and computer graphics, where surfaces are broken into millions of small triangles and each one’s area is found this way.

Why a careers reader should care.

• Surveying and civil engineering: plot areas, cut-and-fill volumes, and parcel boundaries are computed from measured sides and included angles.
• Computer graphics and games: every rendered surface is a mesh of triangles, and lighting, collision, and area calculations run this formula (in its cross-product form) on each one.
• Robotics and navigation: the signed version of $\frac{1}{2}ab\sin C$ tells which way a turn bends, which is how a path-planner decides left from right.

A subtlety worth noticing. Because $\sin C = \sin(180^\circ - C)$, two different triangles on the same two sides, one with an acute included angle and one with the supplementary obtuse angle, have identical areas. The area alone does not pin down the angle. This is the same supplement ambiguity that returns, with consequences, in the law of sines.

Check Yourself

Close the notes and answer each from memory, then reveal it. Pulling an idea back from memory is one of the strongest ways to make it stick.

Resources

Last updated: 2026-06-24