Mathematics · Trigonometry

Triangle Area with Sine

Q: Does the angle really have to lie between the two sides?

Yes, necessarily. In $A = \frac{1}{2}\cdot{}a\cdot{}b\cdot{}\sin{} \gamma{}$, γ is the angle ENCLOSED by a and b; only then is $b\cdot{}\sin{} \gamma{}$ the height on a. If you insert one of the two other angles, you compute the area of a different triangle. Small numerical example: a triangle with $a = 5$, $b = 7$ and included $\gamma{} = 60$° has $A = 17.5\cdot{}\sin{} 60$° ≈ 15.16; wrongly taking an angle of 40° would give ≈ 11.25, clearly off. If only a non-included angle is known, first reduce it to the included one via the angle sum (α + β + γ = 180°) or the law of sines. A labelled sketch shows immediately which angle sits between the given sides.

Q: How do you compute the triangle area from three sides?

With Heron's formula: $A = \sqrt{(s\cdot{}(s - a)\cdot{}(s - b)\cdot{}(s - c))}$, where $s =\frac{ (a + b + c)}{2}$ is the semi-perimeter. Example: sides 5, 6 and 7: $s = 9$, so $A = \sqrt{(9\cdot{}4\cdot{}3\cdot{}2)} = \sqrt{216} \approx{} 14.70$ area units. No angle needed, no height; therefore Heron is the method of choice in the SSS case. Two checks pay off: first, the triangle inequality must hold (each side smaller than the sum of the others), otherwise the radicand is negative and the triangle does not exist. Second, you can verify by computing an angle with the law of cosines and cross-checking with $\frac{1}{2}\cdot{}a\cdot{}b\cdot{}\sin{} \gamma{}$; here: $\cos{} \gamma{} =\frac{ (25 + 36 - 49)}{60} = 0.2$, $\sin{} \gamma{} \approx{} 0.98$, $A = 15\cdot{}0.98 \approx{} 14.7$ ✓.

The trigonometric area formula: two sides and the angle enclosed by them determine the area of the triangle.

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Formula

A = \frac{1}{2} \cdot a \cdot b \cdot sin γ

a and b in cm or m · γ in degrees or radians · A in cm² or m²

Variables & units

Variables & units
Symbol	Meaning	Unit
$A$	Area of the triangle	cm², m²
a, b	Two sides of the triangle	cm, m
$γ$	Angle enclosed by a and b	° or rad

LaTeX

A = \frac{1}{2} \cdot a \cdot b \cdot \sin\gamma

QUANTA STUDYquanta-study.de

Triangle with sides a and b and the included angle γ; the highlighted height onto b is labelled a·sin γ. — The height onto side b is a·sin γ, hence A = ½·b·(a·sin γ) = ½·a·b·sin γ.

Derivation & background

Derivation from $A = \frac{1}{2}\cdot{}g\cdot{}h$ : the height on a is $h = b\cdot{}\sin{} \gamma{}$ , substituting gives the formula. For $\gamma{} = 90$ ° we get $\sin{} \gamma{} = 1$ and $A = \frac{1}{2}ab$ remains, the right triangle. If all three sides are given, Heron's formula $A = \sqrt{(s(s-a)(s-b)(s-c))}$ with $s =\frac{ (a+b+c)}{2}$ helps (Heron of Alexandria, Metrica). In vector form $A = \frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|$ ; the parallelogram has twice the area $ab\sin\gamma$ .

Exam blueprint

Validity range

Holds in every triangle, including obtuse ones; γ must be the angle enclosed by a and b. For $\gamma{} = 90$ ° the formula reduces to A = ½ab.

Derivation steps

The standard formula ½·base·height, with the height from the sine.

1Choose a as the base; the height on it is h = b·sin γ (right partial triangle).
2Substituting into $A = \frac{1}{2}\cdot{}a\cdot{}h$ yields $A = \frac{1}{2}\cdot{}a\cdot{}b\cdot{}\sin{} \gamma{}$ .

Rearrangements

Angle from the area

\sin\gamma = \frac{2A}{a \cdot b}

Caution: γ and 180° − γ have the same sine.

Heron formula (three sides)

A = \sqrt{s(s-a)(s-b)(s-c)}, \quad s = \frac{a+b+c}{2}

When no angle but all sides are known.

Parallelogram

A_{P} = a \cdot b \cdot \sin\gamma

Twice the triangle area; in vector form |a⃗ × b⃗|.

Task variant

$a = 8 cm$ , $b = 6 cm$ , $\gamma{} = 30$ °: compute the triangle area.

$A = \frac{1}{2}\cdot{}8\cdot{}6\cdot{}\sin{} 30$ ° = 24·0.5 = 12 cm².

A triangle has sides 5, 6 and 7. Compute the area with Heron.

$s =\frac{ (5 + 6 + 7)}{2} = 9$ .
$A = \sqrt{(9\cdot{}(9-5)\cdot{}(9-6)\cdot{}(9-7))} = \sqrt{(9\cdot{}4\cdot{}3\cdot{}2)} = \sqrt{216} \approx{} 14.70$ area units.

Common mistakes

Inserting an angle that is not the included one.

γ must lie between a and b; otherwise convert angles first (angle sum).

Forgetting the factor ½.

ab·sin γ is the parallelogram area; the triangle is half of it.

Calculator in the wrong angle mode.

Check $\frac{DEG}{RAD}$ : sin 30° = 0.5, but $\sin{}(30 rad) \approx{} -0.99$ .

Exam context

Triangle and surveying tasks (SAS), areas in vector geometry via the cross product, composite figures.

Formula cluster

Triangle trigonometry

Area formula as the link between law of sines, law of cosines and cross product.

a/sin α = b/sin β = c/sin γ

Law of Sines

c² = a² + b² − 2ab·cos γ

Law of Cosines

a⃗×b⃗ = (a₂b₃−a₃b₂ | a₃b₁−a₁b₃ | a₁b₂−a₂b₁)

Cross Product (Vector Product)

Worked example

$a=5\,\mathrm{cm}$ , $b=7\,\mathrm{cm}$ , $\gamma=60^\circ$ : $A=\frac12\cdot5\cdot7\sin60^\circ=17.5cdot0.866approx15.16\,\mathrm{cm}^2$ . Heron for the sides 5, 6, 7: $s=9$ , $A=\sqrt{9\cdot4\cdot3\cdot2}=\sqrt{216}\approx14.70\,\mathrm{cm}^2$ .

Applications

Triangle calculation in the SAS case, areas of plots and polygons (decomposition into triangles), derivation of the law of sines, parallelogram areas

How do you calculate with Triangle Area with Sine?

Here is how to work through a typical Triangle Area with Sine (A = ½·a·b·sin γ) task step by step:

1
Task
$a = 8 cm$ , $b = 6 cm$ , $\gamma{} = 30$ °: compute the triangle area.
Solution path
$A = \frac{1}{2}\cdot{}8\cdot{}6\cdot{}\sin{} 30$ ° = 24·0.5 = 12 cm².
2
Task
A triangle has sides 5, 6 and 7. Compute the area with Heron.
Solution path
1. $s =\frac{ (5 + 6 + 7)}{2} = 9$ .
2. $A = \sqrt{(9\cdot{}(9-5)\cdot{}(9-6)\cdot{}(9-7))} = \sqrt{(9\cdot{}4\cdot{}3\cdot{}2)} = \sqrt{216} \approx{} 14.70$ area units.

Frequently asked questions about Triangle Area with Sine

When do you use the formula

A = \frac{1}{2}\cdot{}a\cdot{}b\cdot{}\sin{} \gamma{}

Whenever two sides and the INCLUDED angle are given, the classic SAS case. Then the formula delivers the area in one step, without you having to construct a height. Example: $a = 8 cm$ , $b = 6 cm$ , $\gamma{} = 30$ °: $A = \frac{1}{2}\cdot{}8\cdot{}6\cdot{}\sin{} 30$ ° = 24·0.5 = 12 cm². It is also practical when the height is hard to access, e.g. in surveying tasks with measured angles. If the included angle is missing, compute it first via the angle sum or the law of sines; if instead all three sides are given, Heron's formula is the more direct route. For $\gamma{} = 90$ ° the formula turns into the familiar $A = \frac{1}{2}\cdot{}a\cdot{}b$ .

Does the angle really have to lie between the two sides?+

Yes, necessarily. In $A = \frac{1}{2}\cdot{}a\cdot{}b\cdot{}\sin{} \gamma{}$ , γ is the angle ENCLOSED by a and b; only then is $b\cdot{}\sin{} \gamma{}$ the height on a. If you insert one of the two other angles, you compute the area of a different triangle. Small numerical example: a triangle with $a = 5$ , $b = 7$ and included $\gamma{} = 60$ ° has $A = 17.5\cdot{}\sin{} 60$ ° ≈ 15.16; wrongly taking an angle of 40° would give ≈ 11.25, clearly off. If only a non-included angle is known, first reduce it to the included one via the angle sum (α + β + γ = 180°) or the law of sines. A labelled sketch shows immediately which angle sits between the given sides.

How is the sine area formula related to

A = \frac{1}{2}\cdot{}g\cdot{}h

It IS the same formula, just with the height computed. Choose a as the base g. The height h stands perpendicular on a and forms a right partial triangle with side b, in which h is the side opposite the angle γ: $h = b\cdot{}\sin{} \gamma{}$ . Substituting into $A = \frac{1}{2}\cdot{}g\cdot{}h$ immediately gives $A = \frac{1}{2}\cdot{}a\cdot{}b\cdot{}\sin{} \gamma{}$ . So the sine takes over the height construction: instead of measuring or constructing h, you compute it from b and the angle. This also explains the limiting cases: at $\gamma{} = 90$ °, $\sin{} \gamma{} = 1$ and b itself is the height; for very small γ the triangle collapses, sin γ → 0 and the area vanishes.

How do you compute the triangle area from three sides?+

With Heron's formula: $A = \sqrt{(s\cdot{}(s - a)\cdot{}(s - b)\cdot{}(s - c))}$ , where $s =\frac{ (a + b + c)}{2}$ is the semi-perimeter. Example: sides 5, 6 and 7: $s = 9$ , so $A = \sqrt{(9\cdot{}4\cdot{}3\cdot{}2)} = \sqrt{216} \approx{} 14.70$ area units. No angle needed, no height; therefore Heron is the method of choice in the SSS case. Two checks pay off: first, the triangle inequality must hold (each side smaller than the sum of the others), otherwise the radicand is negative and the triangle does not exist. Second, you can verify by computing an angle with the law of cosines and cross-checking with $\frac{1}{2}\cdot{}a\cdot{}b\cdot{}\sin{} \gamma{}$ ; here: $\cos{} \gamma{} =\frac{ (25 + 36 - 49)}{60} = 0.2$ , $\sin{} \gamma{} \approx{} 0.98$ , $A = 15\cdot{}0.98 \approx{} 14.7$ ✓.

Does the formula also work for obtuse angles?+

Yes, without modification. The sine is positive for all angles between 0° and 180°, so $A = \frac{1}{2}\cdot{}a\cdot{}b\cdot{}\sin{} \gamma{}$ yields a positive area even for obtuse γ. Indeed sin(180° − γ) = sin γ: a triangle with $\gamma{} = 120$ ° has the same sine value as one with 60°, namely ≈ 0.866. Example: $a = 5$ , $b = 7$ , $\gamma{} = 120$ °: $A = 17.5\cdot{}0.866 \approx{} 15.16$ area units, exactly as large as for the 60° triangle with the same sides; the two triangles differ but have equal area. Intuitively: for an obtuse angle the height $h = b\cdot{}\sin{} \gamma{}$ falls outside the triangle onto the extension of the base, but keeps its length. Only at $\gamma{} = 0$ ° or 180° does the triangle degenerate and $A = 0$ .

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, Algebra and Trigonometry 2e, §10.2: Non-right Triangles: Law of Cosines

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Symbol

Meaning

Unit

A

Area of the triangle

cm², m²

a, b

Two sides of the triangle

cm, m

γ

Angle enclosed by a and b

° or rad

Derivation & background

Exam blueprint

Validity range

Holds in every triangle, including obtuse ones; γ must be the angle enclosed by a and b. For $\gamma{} = 90$ ° the formula reduces to A = ½ab.

Derivation steps

The standard formula ½·base·height, with the height from the sine.

1Choose a as the base; the height on it is h = b·sin γ (right partial triangle).
2Substituting into $A = \frac{1}{2}\cdot{}a\cdot{}h$ yields $A = \frac{1}{2}\cdot{}a\cdot{}b\cdot{}\sin{} \gamma{}$ .

How do you calculate with Triangle Area with Sine?

Here is how to work through a typical Triangle Area with Sine (A = ½·a·b·sin γ) task step by step:

Task

$a = 8 cm$ , $b = 6 cm$ , $\gamma{} = 30$ °: compute the triangle area.

Solution path

$A = \frac{1}{2}\cdot{}8\cdot{}6\cdot{}\sin{} 30$ ° = 24·0.5 = 12 cm².

Task

A triangle has sides 5, 6 and 7. Compute the area with Heron.

Solution path

$s =\frac{ (5 + 6 + 7)}{2} = 9$ .
$A = \sqrt{(9\cdot{}(9-5)\cdot{}(9-6)\cdot{}(9-7))} = \sqrt{(9\cdot{}4\cdot{}3\cdot{}2)} = \sqrt{216} \approx{} 14.70$ area units.

Frequently asked questions about Triangle Area with Sine

When do you use the formula

A = \frac{1}{2}\cdot{}a\cdot{}b\cdot{}\sin{} \gamma{}

Does the angle really have to lie between the two sides?+

How is the sine area formula related to

A = \frac{1}{2}\cdot{}g\cdot{}h

How do you compute the triangle area from three sides?+

Does the formula also work for obtuse angles?+

Retain Triangle Area with Sine for exams

Create a curated FSRS exam set for A = ½·a·b·sin γ: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential