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Area of a triangle (given the vertical height)

Inverse trig functions

Rounding to decimal places Rounding to significant figures Pythagorean theoremThis topic is relevant for:

Here is everything you need to know about finding the **area of a triangle using trigonometry** for GCSE maths (Edexcel, AQA and OCR). You’ll learn how to generate the area of a triangle formula, use the formula to find the area of a triangle and apply this formula to other polygons.

Look out for the Area of a Triangle worksheets and exam questions at the end.

Area of a triangle trig is a formula to calculate the area of any triangle:

\[\text{Area of triangle }=\frac{1}{2}ab\sin C \]

Previously, we have calculated the area of a triangle using another formula:

\[\text{Area of a triangle }=\frac{\text{base} \times \text{height}}{2}\]

To use this we need to know the vertical height (perpendicular height to the base) of the triangle and the base of the triangle.

We can adapt this formula using the trigonometric ratio \sin(\theta)=\frac{O}{H} to work out the area of a triangle when we do not know its vertical height. The formula we get is:

\[\text{Area of triangle }=\frac{1}{2}ab\sin C\]

The triangle should be labelled as follows, with the lower case letter for each side opposite the corresponding upper case letter for the angle.

We need to know:

- The length of at least
2 sides of the triangle. - The included angle between these two sides.

For example, triangle

If we know or can work out the vertical height of a triangle, it can be easier to use the following formula:

\[\text{Area of a triangle }=\frac{\text{base} \times \text{height}}{2}\]

E.g.

However, if the vertical height is not labelled and we know two sides and the angle in between, we would need to use the following:

\[\text{Area of triangle }=\frac{1}{2}ab\sin C\]

Once we know which formula to use we need to substitute the correct values into it and then solve the equation to calculate the area. The area is always written with square units.

**Remember**: other polygons can be split into triangles to find the interior angles,

so:

\[\text{Area of triangle }=\frac{1}{2}ab\sin C\]

can be applied to find the area of a rectangle, the area of an equilateral triangle, the area of a pentagon, the area of a parallelogram, etc.

**Step by step guide: **Angles in polygons.

In order to find the area of a triangle using

\[\text{Area of triangle }=\frac{1}{2}ab\sin C\]

- Label the angle we are going to use angle
C and its opposite sidec . Label the other two anglesB andA and their corresponding sideb anda . - Substitute the given values into the formula \text{Area }=\frac{1}{2}absinC.
- Solve the equation.

Get your free area of a triangle trig ½abSinC worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free area of a triangle trig ½abSinC worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREECalculate the area of the triangle

**Label the angle we are going to use angle**C and its opposite sidec .**Label the other two angles**A andB and their corresponding sidea andb .

2**Substitute the given values into the formula**

\[A=\frac{1}{2}ab\sin(C)\]

\[A=\frac{1}{2}\times12\times7\times\sin(77)\]

3**Solve the equation.**

\[\begin{aligned}
A&=\frac{81.84708544}{2}\\
A&=40.92\mathrm{cm}^{2}
\end{aligned}\]

Calculate the area of the triangle. Write your answer to

**Label the angle we are going to use angle C and its opposite side c. **

Here, we label each side

**Substitute the given values into the formula**.

\[A=\frac{1}{2}ab\sin(C)\]

\[A=\frac{1}{2}\times18.2\times10\times\sin(65)\]

**Solve the equation.**

\[\begin{aligned}
A&=\frac{164.9480172}{2}\\
A&=82.47\mathrm{m}^{2}
\end{aligned}\]

Calculate the area of the scalene triangle

**Label the angle we are going to use angle C and its opposite side c. **

Here, we label each side and each angle.

** Substitute the given values into the formula**.

\[A=\frac{1}{2}ab\sin(C)\]

\[A=\frac{1}{2}\times5.5\times3.6\times\sin(24)\]

**Solve the equation.**

\[\begin{aligned}
A&=\frac{8.053385533}{2}\\
A&=4.03\mathrm{cm}^{2}
\end{aligned}\]

Triangle

**Label the angle we are going to use angle C and its opposite side c. **

Here, we label each side and each angle.

**Substitute the given values into the formula**.

\[A=\frac{1}{2}ab\sin(C)\]

As the triangle

\[A=\frac{1}{2}\times 9.7\times11.8\times\sin(45)\]

**Solve the equation.**

\[\begin{aligned}
A=\frac{80.93544217}{2}\\
A=40.47\mathrm{km}^{2}
\end{aligned}\]

Calculate the area of the triangle ABC. Write your answer to

Here, we label ^{o} as

**Substitute the given values into the formula**.

\[A=\frac{1}{2}ab\sin(C)\]

Here, we need to be careful to use the correct angle of C=^{o}.

\[A=\frac{1}{2}\times19.1\times15.8\times\sin(22)\]

**Solve the equation.**

\[\begin{aligned}
A&=\frac{113.0487778}{2}\\
A&=56.52\mathrm{cm}^{2}
\end{aligned}\]

Calculate the area of the triangle

**Label each angle (**

Here, we have to think carefully because

\[A=\frac{1}{2}ab\sin(C)\]

As the known sides b and c have the included angle at

\[A=\frac{1}{2}bc\sin(A)\]

**Substitute the given values into the formula**.

\[A=\frac{1}{2}bc\sin(A)\]

\[A=\frac{1}{2}\times 165\times131\times\sin(9)\]

**Solve the equation.**

\[\begin{aligned}
A&=\frac{3381.330962}{2}\\
A&=1691\mathrm{mm}^{2}
\end{aligned}\]

The area of this triangle is ^{2}

** Label each angle (**

**Substitute the given values into the formula**.

\[A=\frac{1}{2}ab\sin(C)\]

This time we know the area, one side and the angle.

Therefore:

\[30=\frac{1}{2}\times x\times12\times\sin(38)\]

**Solve the equation.**

\[\begin{aligned}
30&=6 \times x \times 0.616\\
30&=3.694x\\
8.12&=x
\end{aligned}\]

The area of this triangle is ^{2}

** Label each angle (**

**Substitute the given values into the formula**.

\[A=\frac{1}{2}ab\sin(C)\]

This time we know the area and two sides.

Therefore:

\[42=\frac{1}{2}\times 19\times14 \times\sin(x)\]

**Solve the equation.**

\[\begin{aligned}
42&=133\sin(x)\\
\frac{42}{133}&=\sin(x)\\
0.316&=\sin(x)\\
x&=sin^{-1}(0.316)\\
x&=18.4^{\circ}
\end{aligned}\]

We derive the formula for the area of any triangle by taking the triangle ABC, with vertical height, h:

By applying the usual formula for the area of a triangle (\frac{\text{base}\times\text{height}}{2}) we have A=\frac{1}{2}(a\times{h}).

We can also state, using trigonometry, that \sin(C)=\frac{h}{b} which we can rearrange to make h the subject h=b\sin(C).

Substituting h=b\sin(C) into A=\frac{1}{2}(a\times{h}), we obtain: A=\frac{1}{2}ab\sin(C).

It is important to notice that C is the included angle between the sides of a and b.

**Incorrectly labelling the triangle so the substitution is incorrect**

**Not using the included angle between a and b.**

**The triangle is assumed to contain a right angle and so the area is calculated by halving the base times the height.**

**Using inverse sine instead of sine of the angle to find the area.**

Area of a triangle trig is part of our series of lessons to support revision on trigonometry. You may find it helpful to start with the main trigonometry lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

1. Calculate the area of the right angle triangle.

189\mathrm{m}^{2}

163.7\mathrm{m}^{2}

94.5\mathrm{m}^{2}

378\mathrm{m}^{2}

Label the triangle:

\begin{aligned} \text{Area }&=\frac{1}{2}ab \sin(C)\\ \text{Area }&=\frac{1}{2} \times 21 \times 18 \times \sin(30)\\ \text{Area }&=94.5 \mathrm{m}^{2} \end{aligned}

2. Calculate the area of the triangle, correct to 2 decimal places.

32.85\mathrm{m}^{2}

7.95\mathrm{m}^{2}

31.87\mathrm{m}^{2}

18.60\mathrm{m}^{2}

Label the triangle:

\begin{aligned} \text{Area }&=\frac{1}{2}ab \sin(C)\\ \text{Area }&=\frac{1}{2} \times 7.3 \times 9 \times \sin(76)\\ \text{Area }&=31.87 \mathrm{m}^{2} \end{aligned}

3. Calculate the area of the equilateral triangle XYZ. Write your answer to 2 decimal places.

74.31\mathrm{m}^{2}

42.90\mathrm{m}^{2}

85.81\mathrm{m}^{2}

81.72\mathrm{m}^{2}

Label the triangle:

\begin{aligned} \text{Area }&=\frac{1}{2}ab \sin(C)\\ \text{Area }&=\frac{1}{2} \times 13.1 \times 13.1 \times \sin(60)\\ \text{Area }&=74.31 \mathrm{m}^{2} \end{aligned}4. Calculate the area of the parallelogram, correct to 2 decimal places.

125.41\mathrm{mm}^{2}

250.81\mathrm{mm}^{2}

376.22\mathrm{mm}^{2}

225.83\mathrm{mm}^{2}

We need to look at the two triangles individually. The triangles are congruent (exactly the same) since all three of their lengths are equal (SSS). Therefore we can calculate the area of one triangle and then double it.

Label one triangle:

\begin{aligned} \text{Area }&=\frac{1}{2}ab \sin(C)\\ \text{Area }&=\frac{1}{2} \times 15 \times 22.5 \times \sin(48)\\ \text{Area }&=125.406 \mathrm{mm}^{2}\\ \text{Total area }&=2\times125.406=250.81\mathrm{mm}^{2} \end{aligned}

5. Calculate the area of the isosceles triangle PQR, correct to 3 significant figures.

18.7\mathrm{cm}^{2}

12.4\mathrm{cm}^{2}

10.6\mathrm{cm}^{2}

17.3\mathrm{cm}^{2}

First, calculate angle PQR: 180-28.3-28.3=123.4{\circ} .

Then label the triangle:

\begin{aligned} \text{Area }&=\frac{1}{2}ab \sin(C)\\ \text{Area }&=\frac{1}{2} \times 6.7 \times 6.7 \times \sin(123.4)\\ \text{Area }&=18.7 \mathrm{cm}^{2} \end{aligned}6. Calculate the value of \theta

\theta=60^{\circ}

\theta=14.5^{\circ}

\theta=0.5^{\circ}

\theta=30^{\circ}

\begin{aligned}
\text{Area } &=\frac{1}{2}ab \sin(C)\\
210&=\frac{1}{2} \times 24 \times 35 \times \sin(\theta)\\
210&=420 \sin(\theta)\\
0.5&= \sin(\theta)\\
\sin^{-1}(0.5)&=\theta\\
30^{\circ}&=\theta
\end{aligned}

1. In triangle {katex]ABC[/katex], AB=8m, AC=18m and angle BAC=31^{\circ}. Calculate the area of triangle ABC.

**(2 marks)**

Show answer

A=\frac{1}{2}\times 8 \times 18 \times \sin(31)

**(1)**

**(1)**

2. Quadrilateral ABCD is made from two triangles.

a) Work out the length AC.

b) Calculate the total area of the quadrilateral.

**(5 marks)**

Show answer

\begin{aligned}
11^{2}&=7^{2}+b^{2}\\
b^{2}&=11^{2}-7^{2}
\end{aligned}

**(1)**

**(1)**

\text{Area ABC: } \frac{1}{2} \times 7 \times 8.49 =29.72 \mathrm{cm}^{2}

**(1)**

**(1)**

**(1)**

3. The area of triangle PQR is 55cm^2. Work out the value of x. Give your answer to 2dp.

**(4 marks)**

Show answer

\frac{1}{2} \times x \times 2x \times \sin(29) = 55

**(1)**

**(1)**

**(1)**

**(1)**

You have now learned how to:

- Know and apply \text{Area}=\frac{1}{2}ab\sin C to calculate the area, sides or angles of any triangle

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