We know how to find the area of right-angled triangles - we just multiply the two short sides together, and halve the result:
The area of this triangle is $\frac{1}{2}ab$12ab, half the area of the rectangle.
If the triangle is not right-angled we can still find the area, as long as we know two sides and the angle between them:
The area of this triangle is the base $a$a times the height, and then halved, just like for right-angled triangles. But what is the height? It isn't $b$b in this case, but we can use $b$b and the angle $C$C to find it.
Here we have made a small right-angled triangle within our larger triangle, with hypotenuse $b$b and short side $h$h, the height of our large triangle. Using trigonometric ratios, the value of $\sin C$sinC is the opposite side, $h$h, divided by the hypotenuse, $b$b.
So $\sin C=\frac{h}{b}$sinC=hb and hence, $h=b\sin C$h=bsinC.
Putting this all together with the area formula $Area=\frac{1}{2}base\times height$Area=12base×height, we obtain the formula:
If a triangle has sides of length $a$a and $b$b, and the angle between these sides is $C$C, then
$\text{Area }=\frac{1}{2}ab\sin C$Area =12absinC
Practice questions
Question 1
Calculate the area of the following triangle.
Round your answer to two decimal places.
Question 2
$\triangle ABC$△ABC has an area of $520$520 cm2. The side $BC=48$BC=48 cm and $\angle ACB=35^\circ$∠ACB=35°.
What is the length of $b$b?
Round your answer to the nearest centimetre.
Given $\triangle ABC$△ABC with length of side $AC$AC labeled as $b$b cm and side $BC$BC labeled as $48$48 cm and their interior $\angle ACB$∠ACB that measures $35$35º. $\angle ACB$∠ACB is an included angle of the two given sides. Both side AC and BC are adjacent to the given included angle.
question 3
Using the triangle shown below, solve for the possible value(s) of $x$x.
If there is more than one answer, write all answers on the same line, separated by commas. Round your answer(s) to one decimal place.
The diagram is not given to scale.
Further applications
Many shapes can be divided into triangles and if we have sufficient information we can apply the sine area rule to solve problems involving areas.
Worked examples
Example 1
Find the area of a regular pentagon inscribed in a circle of radius $8$8 cm.
Pictured left is a regular pentagon inscribed in a circle. Drawing lines out from the centre to the corners of the pentagon we will form 5 congruent triangles. Each triangle will be isosceles with two sides equal to the radius. The central angle can be found by dividing a full rotation by 5:
| $\theta$θ | $=$= | $\frac{360^\circ}{5}$360°5 |
| $=$= | $72^\circ$72° |
We now have enough information to use our area rule.
| Area | $=$= | $5\times\frac{1}{2}ab\sin C$5×12absinC |
| $=$= | $5\times\frac{1}{2}\times8\times8\times\sin\left(72^\circ\right)$5×12×8×8×sin(72°) | |
| $=$= | $152$152$cm^2$cm2 (to the nearest $cm^2$cm2) |
Example 2
A rhombus has area $50$50$cm^2$cm2 and its acute angle is $30^\circ$30°. What is the side length of the rhombus?
A rhombus can be split into two isosceles triangles with side length $x$x cm and angle between the sides $30^\circ$30°.
Hence, the area could be found using the formula:
| Area | $=$= | $2\times\frac{1}{2}ab\sin C$2×12absinC |
| $=$= | $x^2\times\sin\left(30^\circ\right)$x2×sin(30°) |
Substituting our known area and rearrange for $x$x:
| $50$50 | $=$= | $x^2\times\sin\left(30^\circ\right)$x2×sin(30°) |
| $50$50 | $=$= | $x^2\times\frac{1}{2}$x2×12 |
| $100$100 | $=$= | $x^2$x2 |
| $x$x | $=$= | $10$10, since $x$x is positive. |
The side length of the rhombus is $10$10 cm.