How do you find a side of a triangle given its area?

Substitute the area, the known side and the included angle into Area equals one half a b sin C, then rearrange for the unknown side: b equals two times the area divided by a sin C. For example if the area is 30, one side is 10 and the included angle is 60 degrees, then b equals 60 divided by 10 sin 60 degrees, which is about 6.93.

When do you combine the area formula with the sine or cosine rule?

When the two sides and the included angle are not all given directly. Find the missing side with the sine rule, or a missing angle from three sides with the cosine rule, and then apply the area formula with two sides and their included angle.

How do you find a side of a triangle given its area?

Substitute the area, the known side and the included angle into Area equals one half a b sin C, then rearrange for the unknown side: b equals two times the area divided by a sin C. For example if the area is 30, one side is 10 and the included angle is 60 degrees, then b equals 60 divided by 10 sin 60 degrees, which is about 6.93.

When do you combine the area formula with the sine or cosine rule?

When the two sides and the included angle are not all given directly. Find the missing side with the sine rule, or a missing angle from three sides with the cosine rule, and then apply the area formula with two sides and their included angle.

Year 12 Methods (Unit 3 & 4) Trigonometry using the sine and cosine rules

The area of a triangle

Q: What is the area of a triangle formula in trigonometry?

For any triangle the area is one half times a times b times the sine of C, where a and b are two sides and C is the angle included between them. It needs two sides and the angle between them, not a perpendicular height, so it works for non-right-angled triangles.

Q: What is the included angle?

The included angle is the angle enclosed by the two sides you use in the formula, that is the angle at the vertex where those two sides meet. Using an angle that is not between the two chosen sides gives the wrong area.

Q: How do you find the included angle given the area?

Rearrange to sin C equals two times the area divided by a b, then take the inverse sine. Because sine of C equals sine of 180 degrees minus C, there is both an acute and an obtuse answer in a triangle, and both give the same area.

Q: Is Heron's formula needed in QCAA Methods?

No. The Queensland Mathematical Methods syllabus lists Area equals one half b c sin A. When only the three sides are known, find an angle with the cosine rule first, then use the area formula.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the area of any triangle is $\text{Area}=\dfrac{1}{2}\,a\,b\,\sin C$, where $a$ and $b$ are two sides and $C$ is the included angle between them. Rearranged, it finds an unknown side or the included angle from a known area; combined with the sine and cosine rules, it handles triangles where a part must be found first.

The trigonometric area formula works for any triangle. If you know two sides and the angle enclosed by them, the area is $\dfrac{1}{2}\,a\,b\,\sin C$. Because it uses the sine of the included angle instead of a perpendicular height, it applies to non-right-angled triangles where $\tfrac12\times\text{base}\times\text{height}$ would be awkward.

The included angle is the one at the vertex where the two chosen sides meet. In $\dfrac{1}{2}\,a\,b\,\sin C$, the side $a$, the side $b$ and the angle $C$ form a matched set: $C$ sits between $a$ and $b$. Choosing an angle that is not between the two sides is the most common error.

The same formula, rearranged, finds a missing side ($b=\dfrac{2\times\text{Area}}{a\sin C}$) or the missing included angle ($\sin C=\dfrac{2\times\text{Area}}{a\,b}$) when the area is given. If the two sides and their included angle are not all supplied, use the sine rule or cosine rule to find the missing part first. The formula is central to modelling triangular regions in land, gardens and design.

Key idea. \(\text{Area}=\dfrac{1}{2}\,a\,b\,\sin C\): pick two sides and the angle between them. Rearrange for a side, or for \(\sin C\) to get the included angle.

Two sides $a,b$ and the included angle $C$ between them

A given area yields two included angles: an acute one and its obtuse supplement

The area of a triangle from two sides and the included angle:

\[\text{Area}=\frac{1}{2}\,a\,b\,\sin C\]

Area = \frac{1}{2} a b \sin C

Rearranged to find a side, and to find the included angle:

\[b=\frac{2\,\text{Area}}{a\,\sin C}\qquad\qquad \sin C=\frac{2\,\text{Area}}{a\,b}\]

b = \frac{2 Area}{a \sin C}

Supporting rules for two-step problems (find a missing part first):

\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\qquad\qquad c^2=a^2+b^2-2ab\cos C\]

c^{2} = a^{2} + b^{2} - 2 a b \cos C

Two solutions. When you solve \(\sin C=k\) for the included angle, both the acute value and its obtuse supplement \(180^\circ-C\) give the same area.

How to use the area formula

Identify the parts. You need two sides and the angle between them. Sketch and label the triangle so the included angle is clear.
Finding the area? Substitute straight into $\text{Area}=\dfrac{1}{2}\,a\,b\,\sin C$. Make sure the calculator is in the right mode (degrees or radians as given).
Finding a side? Substitute the area, the known side and the included angle, then rearrange: $b=\dfrac{2\times\text{Area}}{a\sin C}$.
Finding the included angle? Rearrange to $\sin C=\dfrac{2\times\text{Area}}{a\,b}$, take the inverse sine, and give both the acute answer and its obtuse supplement $180^\circ-C$ if the question allows either.
Missing a part? Use the sine rule (a side and its opposite angle) or the cosine rule (three sides, or two sides and the included angle) to find it first, then apply the area formula.

Rounding. Keep at least four figures inside the working and round only the final answer; give areas to the stated precision and angles to the nearest degree unless told otherwise.

Example 1 — direct area

A triangle has sides $9$ cm and $12$ cm with an included angle of $40^\circ$. Find its area, to one decimal place.

Solution

Two sides and the angle between them.

Area	$=$	$\dfrac{1}{2}\times 9\times 12\times \sin 40^\circ$
	$=$	$54\sin 40^\circ$
	$=$	$34.7\ \text{cm}^2$

Area = 34.7

Example 2 — find a side

A triangle of area $30$ cm$^2$ has a $10$ cm side enclosing a $60^\circ$ angle. Find the other enclosing side.

Solution

Substitute, then rearrange for $b$.

$30$	$=$	$\dfrac{1}{2}\times 10\times b\times \sin 60^\circ$
$30$	$=$	$4.330\,b$
$b$	$=$	$6.93\ \text{cm}$

b = 6.93

Example 3 — find the included angle

A triangle with sides $8$ and $11$ has an area of $40$ cm$^2$. Find the possible size(s) of the included angle.

Solution

Rearrange for $\sin C$; there are two answers.

$\sin C$	$=$	$\dfrac{2\times 40}{8\times 11}=0.9091$
$C$	$=$	$65.4^\circ\ \text{or}\ 114.6^\circ$

Both give area $40$ cm$^2$, since $\sin C=\sin(180^\circ-C)$.

C = 65.4 ° or 114.6 °

Example 4 — applied (modelling)

A triangular block of land has two boundaries of $42$ m and $35$ m meeting at $108^\circ$. Find its area, and the cost of turf at $\$18$ per square metre.

Solution

The two boundaries are the sides; $108^\circ$ is included.

Area	$=$	$\dfrac{1}{2}\times 42\times 35\times \sin 108^\circ$
	$=$	$699\ \text{m}^2$
cost	$=$	$699\times \$18=\$12{,}582$

Area = 699

Common pitfalls

Use the included angle. The angle must be the one between the two sides you put into the formula. A non-included angle gives the wrong area.

Keep the $\tfrac12$, and use sine. Forgetting the $\tfrac12$ doubles the area; using $\cos$ instead of $\sin$ is a common slip.

Two angles for one area. When you find the included angle from a given area, remember the obtuse supplement $180^\circ-C$ gives the same area — state both if the question allows it.

Only three sides? Heron's formula is not required in QCAA Methods — find an angle with the cosine rule first, then use $\tfrac12 a b\sin C$.

Frequently asked questions

What is the area of a triangle formula in trigonometry?

$\text{Area}=\dfrac{1}{2}\,a\,b\,\sin C$, where $a$ and $b$ are two sides and $C$ is the angle included between them. It works for any triangle.

What is the included angle?

The angle at the vertex where the two chosen sides meet — the angle between $a$ and $b$. Using any other angle gives the wrong area.

How do you find a side given the area?

Rearrange to $b=\dfrac{2\times\text{Area}}{a\sin C}$. For area $30$, side $10$ and included angle $60^\circ$, $b=\dfrac{60}{10\sin 60^\circ}\approx 6.93$.

How do you find the included angle given the area?

Use $\sin C=\dfrac{2\times\text{Area}}{a\,b}$, then take the inverse sine. Both the acute value and its obtuse supplement $180^\circ-C$ give the same area.

When do you combine it with the sine or cosine rule?

When the two sides and their included angle are not all given. Find the missing part with the sine rule (a side and its opposite angle) or the cosine rule (three sides), then apply the area formula.

Is Heron's formula needed in QCAA Methods?

No. The syllabus lists $\text{Area}=\tfrac12 b c\sin A$. With only three sides, find an angle by the cosine rule first, then use the area formula.

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More in Trigonometry using the sine and cosine rules

Area	\(=\)	\(\dfrac{1}{2}\times 9\times 12\times \sin 40^\circ\)
	\(=\)	\(54\sin 40^\circ\)
	\(=\)	\(34.7\ \text{cm}^2\)

\(30\)	\(=\)	\(\dfrac{1}{2}\times 10\times b\times \sin 60^\circ\)
\(30\)	\(=\)	\(4.330\,b\)
\(b\)	\(=\)	\(6.93\ \text{cm}\)

\(\sin C\)	\(=\)	\(\dfrac{2\times 40}{8\times 11}=0.9091\)
\(C\)	\(=\)	\(65.4^\circ\ \text{or}\ 114.6^\circ\)

Area	\(=\)	\(\dfrac{1}{2}\times 42\times 35\times \sin 108^\circ\)
	\(=\)	\(699\ \text{m}^2\)
cost	\(=\)	\(699\times \$18=\$12{,}582\)