---
title: "Triangle Area with Half ab sin C"
description: "Find the area of any triangle from two sides and the angle between them, and run the same rule backwards to find an angle from a known area."
canonical: https://lightmysky.com/learn/mathematics/triangle-area-with-half-ab-sin-c-mt_PiwizT4D2b
source: https://lightmysky.com/learn/mathematics/triangle-area-with-half-ab-sin-c-mt_PiwizT4D2b.md
retrieved: 2026-11-12
---

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# Triangle Area with Half ab sin C

Find the area of any triangle from two sides and the angle between them, and run the same rule backwards to find an angle from a known area.

Subject: Mathematics · Area: Geometry · Ages 15 to 16
Page: https://lightmysky.com/learn/mathematics/triangle-area-with-half-ab-sin-c-mt_PiwizT4D2b

## Ready when they can

- Use the two sides either side of the given angle, not any two sides
- Find an angle when the area and two sides are known
- Split a quadrilateral into two triangles to find its area

## Lesson: Half, times, times sine

Half base times height always works for area, but you need the height. When nobody hands you a height, build one. Take a triangle with sides a and b and the angle C between them. Use side a as the base. Drop a perpendicular from the far corner to the base, making a right triangle whose hypotenuse is b and whose height is h. In that right triangle, h equals b sin C. Substitute into the base-height formula and the area is one half times a times b times sin C. The rule is that a and b must touch angle C: grab a far side and the answer breaks.

*(drawing: Drop a perpendicular from the far corner to the base, then read the height from the right triangle.)*

When angle C is obtuse, the perpendicular meets the extension of the base instead of the base itself. The auxiliary triangle still has hypotenuse b and an acute angle of 180 minus C, and sin of 180 minus C equals sin C, so the height is unchanged. At exactly 90 degrees the perpendicular is side b itself, which is the right-angle case of the same formula.

The formula also runs backwards. When you know the area and two sides, solve for the sine of the angle between them: sin C equals 2 times area over a times b, then take the inverse sine. Sine only reaches 1, so the area can never exceed half of a times b, and that maximum happens only at 90 degrees. When sin C is exactly 1 both candidates coincide at 90 degrees, counted once. For any smaller sine, both C and 180 minus C are possible, check which one fits your triangle.

**Example.** A four-sided shape needs one extra move. Draw a diagonal that stays inside the shape to split the quadrilateral into two triangles, find each triangle's area with the same formula, then add the two areas together. A split giving 12 and 7 makes 19 for the whole shape, and the same adding works whatever each half measures. In a concave quadrilateral one diagonal can fall outside, so always choose the internal one.

**Recap.** Multiply the two touching sides by the sine between them, halve it, and split quadrilaterals before adding.

## Practice

16 questions on this page, each with its working shown.

## Needs first

- [The Cosine Rule](https://lightmysky.com/learn/mathematics/the-cosine-rule-mt_yy73gK5CEP)

## Opens up

- [Pythagoras and Trigonometry in Three Dimensions](https://lightmysky.com/learn/mathematics/pythagoras-and-trigonometry-in-three-dimensions-mt_7yKxYqQHB2)
- [Determinants and What They Measure](https://lightmysky.com/learn/mathematics/determinants-and-what-they-measure-mt_fWr2D_1BdM)
- [The Cross Product and Oriented Area](https://lightmysky.com/learn/mathematics/the-cross-product-and-oriented-area-mt_QEBM_Ms-SK)
- [Arc Length, Sector Area and Segments](https://lightmysky.com/learn/mathematics/arc-length-sector-area-and-segments-mt_z9LIwmfKYJ)
