Trigonometry · Ordinary Level
Area of a Triangle
Ordinary Level · Class 10 · Tap NEXT to begin
Section 1 of 2
Area of a Triangle
Two sides and the angle between
$\text{Area} = \tfrac{1}{2}ab\sin C$ — half, times two sides, times the sine of the angle in between them. The answer is in square units.
Worked example — straightforward area
Find the area of a triangle with sides $5$ and $6$ and an angle of $60^\circ$ between them.
$\text{Area} = \tfrac{1}{2}(5)(6)\sin 60 = 12.99 = 13\,\text{cm}^2$
$13\,\text{cm}^2$
Worked example — another
Find the area of a triangle with sides $12\,\text{m}$ and $8\,\text{m}$ and an angle of $75^\circ$ between them.
$\text{Area} = \tfrac{1}{2}(12)(8)\sin 75 = 46.36 = 46.4\,\text{m}^2$
$46.4\,\text{m}^2$
Worked example — working backwards
A triangle with sides $5$ and $6$ has area $7.5\,\text{m}^2$. Find the angle $C$ between them.
$7.5 = \tfrac{1}{2}(5)(6)\sin C = 15\sin C$
$\sin C = \dfrac{7.5}{15} = 0.5 \Rightarrow C = 30^\circ$
$C = 30^\circ$
Section 2 of 2
Find the Angle First
Need the in-between angle
If the given angles are not the one between your two sides, find it first: the three angles add to $180^\circ$.
Worked example — included angle then area
A triangle has angles $23^\circ$ and $98^\circ$, with sides $5\,\text{m}$ and $7\,\text{m}$ meeting at the third vertex. Find the area.
Third angle $= 180 - (23 + 98) = 59^\circ$ (this is the angle between the $5$ and $7$)
$\text{Area} = \tfrac{1}{2}(5)(7)\sin 59 = 15\,\text{m}^2$
$15\,\text{m}^2$
You try
In $\triangle PQR$, $|PQ| = 8$, $|QR| = 7$, $|PR| = 9$. You found $|\angle PQR| = 73^\circ$. Find the area.
$\text{Area} = \tfrac{1}{2}(8)(7)\sin 73$.
$\text{Area} = \tfrac{1}{2}(8)(7)\sin 73 = 26.77 = 26.8\,\text{m}^2$
$26.8\,\text{m}^2$
You try
In $\triangle ABC$, $|AB| = 6$, $|BC| = 5$, $|\angle BAC| = 50^\circ$. Find (i) $|\angle ACB|$, then (ii) the area.
Sine Rule for $C$: $\dfrac{\sin C}{6} = \dfrac{\sin 50}{5}$; then third angle; then area with two sides and that angle.
(i) $\sin C = \dfrac{6\sin 50}{5} = 0.919 \Rightarrow C = 67^\circ$
third angle $= 180 - (50 + 67) = 63^\circ$
(ii) $\text{Area} = \tfrac{1}{2}(5)(6)\sin 63 = 13.36 = 13.4\,\text{cm}^2$
$C = 67^\circ,\ \text{Area} = 13.4\,\text{cm}^2$
You try
In $\triangle PQR$, $|PQ| = 10$, $|PR| = 8$, $|QR| = 9$. Find (i) $|\angle PQR|$, then (ii) the area.
The angle at $Q$ faces $|PR| = 8$: $8^2 = 10^2 + 9^2 - 2(10)(9)\cos Q$. Then area with the two sides meeting at $Q$.
$8^2 = 10^2 + 9^2 - 2(10)(9)\cos Q \Rightarrow 64 = 181 - 180\cos Q$
$\cos Q = \dfrac{117}{180} \Rightarrow Q = 49.5^\circ$
$\text{Area} = \tfrac{1}{2}(10)(9)\sin 49.5 = 34.2\,\text{m}^2$
$Q = 49.5^\circ,\ \text{Area} = 34.2\,\text{m}^2$
That’s Class 10.
Area $= \tfrac{1}{2}ab\sin C$, backwards for an angle, and finding the in-between angle first. Class 11: full exam-style problems.