Area of a Triangle from Two Sides and an Angle

Quick answer

One half base times height needs the height, which is often not given. When two sides and the angle between them are known, the height can be found with sine: drop a perpendicular from a vertex, and the height is a sin C. That gives Area = ½ab sin C. The formula holds for obtuse angles too, because an angle and its supplement have the same sine.

What you'll learn

Derive Area = ½ab sin C by drawing an altitude
Explain why the formula works for obtuse angles
Find the area of a triangle from two sides and the included angle

When the height is missing

A triangle has sides $a = 6$ and $b = 8$ , and the angle between them is $C = 50°$ . What is its area?

One half base times height is the formula to reach for, and it stalls at once: the base can be $b = 8$ , but no height is given. The height has to come from the angle.

Drawing the height

Put the triangle with $C$ at the origin and side $b$ along the $x$ -axis. From vertex $B$ , drop a perpendicular to side $CA$ . Call its foot $D$ and its length $h$ .

The height from B makes a right triangle

Triangle $BDC$ has a right angle at $D$ . Its hypotenuse is $CB = a$ , and the side opposite angle $C$ is the height. So by the sine ratio:

\sin C = \frac{h}{a} \qquad\Longrightarrow\qquad h = a \sin C

Now the base-times-height formula has everything it needs:

\text{Area} = \tfrac{1}{2} \cdot b \cdot h = \tfrac{1}{2} \cdot b \cdot a \sin C

\boxed{\text{Area} = \tfrac{1}{2}ab\sin C}

Two sides and the angle between them are enough to find the area. The angle has to be the one between the two sides, called the included angle.

Why the formula survives an obtuse angle

If $C$ is obtuse, the perpendicular from $B$ misses side $CA$ altogether and lands on its extension, beyond $C$ .

With an obtuse angle at C, the height lands outside

The right triangle $BDC$ still exists, but its angle at $C$ is now the outside angle, $180° - C$ . So the height is

h = a \sin(180° - C)

On the unit circle, an angle and its supplement reach the same height: $\sin(180° - C) = \sin C$ . The height is $a \sin C$ after all, and the formula comes out unchanged.

For a right angle, $\sin 90° = 1$ and the formula becomes $\tfrac{1}{2}ab$ , the familiar area of a right triangle with legs $a$ and $b$ . One formula covers acute, right and obtuse triangles.

Choosing a formula

What is known	Area
a base and its height	$\tfrac{1}{2}bh$
two sides and the included angle	$\tfrac{1}{2}ab\sin C$
two legs of a right triangle	$\tfrac{1}{2}ab$ , the case $C = 90°$

The letters are not fixed. With sides $b$ and $c$ and the angle $A$ between them, the area is $\tfrac{1}{2}bc\sin A$ .

Worked examples

Common mistakes

Practice problems

Find the area of a triangle with sides $10$ and $7$ and an included angle of $30°$ .

Answer

$17.5$

Full solution

$\tfrac{1}{2}(10)(7)\sin 30° = 35 \cdot \tfrac{1}{2} = 17.5$ .
Find the area of a triangle with sides $12$ and $9$ and an included angle of $90°$ .

Answer

$54$

Full solution

$\sin 90° = 1$ , so the area is $\tfrac{1}{2}(12)(9) = 54$ , the right-triangle formula.
Find the area of a triangle with sides $5$ and $8$ and an included angle of $150°$ .

Answer

$10$

Full solution

$\sin 150° = \sin 30° = \tfrac{1}{2}$ , so the area is $\tfrac{1}{2}(5)(8)\cdot\tfrac{1}{2} = 10$ .
Find the exact area of an equilateral triangle with side $6$ .

Answer

$9\sqrt{3}$

Full solution

Every angle is $60°$ , so the area is $\tfrac{1}{2}(6)(6)\sin 60° = 18 \cdot \tfrac{\sqrt{3}}{2} = 9\sqrt{3}$ .
Find the area of a triangle with sides $9$ and $4$ and an included angle of $110°$ , to two decimal places.

Answer

About $16.91$

Full solution

$\tfrac{1}{2}(9)(4)\sin 110° = 18 \sin 110° \approx 18(0.9397) \approx 16.91$ .
A triangle with sides $8$ and $10$ has area $20$ and an acute included angle. Find the angle.

Answer

$30°$

Full solution

$20 = 40\sin C$ gives $\sin C = \tfrac{1}{2}$ . The acute angle with that sine is $30°$ .
Explain why the formula becomes $\tfrac{1}{2}ab$ when $C = 90°$ .

Answer

$\sin 90° = 1$ .

Full solution

With a right angle between $a$ and $b$ , each side is the height on the other. The formula agrees: $\tfrac{1}{2}ab\sin 90° = \tfrac{1}{2}ab$ .
Find the area of a parallelogram with sides $12$ and $5$ and an angle of $30°$ between them.

Answer

$30$

Full solution

Two congruent triangles, each $\tfrac{1}{2}(12)(5)\sin 30° = 15$ , give $30$ .
Two sides of a triangle are $7$ and $7$ . Which included angle gives the largest area, and what is that area?

Hint

Which angle has the largest sine?

Answer

$90°$ , giving $24.5$ .

Full solution

The area is $\tfrac{1}{2}(7)(7)\sin C = 24.5\sin C$ . Sine is largest, $1$ , at $90°$ , so the largest area is $24.5$ .

Opening the angle past $90°$ lowers the height again, which is why the area shrinks on both sides of a right angle.
A triangle has sides $7$ and $9$ , and the angle opposite the side of length $9$ is $60°$ . Jon computes the area as $\tfrac{1}{2}(7)(9)\sin 60°$ . Find his error.

Hint

Is the $60°$ angle between the sides $7$ and $9$ ?

Answer

The $60°$ angle is not between the two known sides, so the formula does not apply to it.

Full solution

The formula comes from a height drawn on the included angle. Here the $60°$ angle sits across from the side of length $9$ , so it is not the angle between $7$ and $9$ .

Jon needs the included angle first. The Law of Sines finds the angle opposite $7$ , then the angle sum gives the included angle, and only then does $\tfrac{1}{2}ab\sin C$ apply.

Frequently asked questions

What is the formula for the area of a triangle with two sides and an angle?

Area = ½ab sin C, where a and b are two sides and C is the angle between them.

Why does the formula use sine?

The height from one vertex to the opposite side is a sin C, from the right triangle that the height creates. Sine turns a side and an angle into a height.

Does it work if the angle is obtuse?

Yes. The height falls outside the triangle, but it equals a sin(180° − C), and sin(180° − C) = sin C.

Which angle do I use?

The angle between the two known sides, called the included angle. An angle opposite one of them does not work in this formula.

How does it relate to one half base times height?

It is the same formula with the height written as a sin C. When C is 90°, sin C = 1 and it becomes ½ab, the right-triangle area.

What to learn next

The Law of Sines, and the Ambiguous CaseWhy each side of a triangle over the sine of its opposite angle gives the same number, how to solve triangles with it, and why SSA can give two triangles or none.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSG.SRT.D.9Similarity, Right Triangles, and Trigonometry(+) Derive the formula A = 1/2 ab sin(C) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side.

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