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Area of a triangle

Area of a triangle

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Summary

Area of a triangle

In a nutshell

The trigonometric area of a triangle can be used to find the area of a triangle without being given the perpendicular height.



Area of a triangle formula

The trigonometric formula for the area of a triangle is used when you have two sides and an angle between them. This is sufficient information to find the area. The area is given to be:


​Area=12absin⁡(C)Area=\dfrac{1}{2}ab\sin(C)Area=21​absin(C)​​


This is for a triangle with sides a,b,ca,b,ca,b,c​ and corresponding angles A,B,CA,B,CA,B,C.


Maths; Trigonometry; KS5 Year 12; Area of a triangle


Example 1

A triangle has two sides with lengths 2cm2cm2cm and 4cm4cm4cm​​ and the angle between them is 24∘24^\circ24∘. Find the area of the triangle to two decimal places.


Sketch and label the sides and angles of this triangle first.


Maths; Trigonometry; KS5 Year 12; Area of a triangle


Now, use the formula:

​Area=12absin⁡(C)Area=\dfrac{1}{2}ab\sin(C)Area=21​absin(C)​​


Substitute in known values:

​Area=12(2)(4)sin⁡(24)Area = \dfrac{1}{2}(2)(4)\sin(24)Area=21​(2)(4)sin(24)​​


​Area=4×sin⁡(24)=1.62694....Area =4\times \sin(24)=1.62694....Area=4×sin(24)=1.62694....​​


​Area =1.63cm2 (2 d.p.)‾\underline{Area\,=1.63cm^2 \space (2\space d.p.)}Area=1.63cm2 (2 d.p.)​


Note: If a triangle is given as a sketch, the labels will not necessarily match the variables in the formula for the area of a triangle. In this case it is possible to adjust the formula to fit the situation.


Example 2

In △ABC\triangle ABC△ABC, BC=(x+3) cmBC= (x+3) \ cmBC=(x+3) cm, AC=x cmAC = x \ cmAC=x cm and ∠BCA=130∘\angle {BCA} = 130^{\circ} ∠BCA=130∘. The area of this triangle is 6cm26cm^26cm2. What is the value of xxx to two decimal places?


​∠BCA\angle BCA∠BCA is between sides ACAC AC and BCBCBC. 

Use the formula for area:

Area=12absin⁡(C)Area=\dfrac{1}{2}ab\sin(C)Area=21​absin(C)​​


Substitute in known values:

6=12(x+3)(x)sin⁡(130)6 = \dfrac12(x+3)(x)\sin(130)6=21​(x+3)(x)sin(130)​​


Solve for xxx:

6=12(x+3)(x)sin⁡(130)12=(x+3)(x)sin⁡(130)12sin⁡(130)=(x2+3x)0=x2+3x−12sin⁡(130)\begin{aligned}6 &=\dfrac12(x+3)(x)\sin(130) \\ \\ 12 &= (x+3)(x)\sin(130) \\ \\ \dfrac{12}{\sin(130)} &= (x^2+3x) \\ \\ 0 &= x^2+3x - \dfrac{12}{\sin(130)} \end{aligned}612sin(130)12​0​=21​(x+3)(x)sin(130)=(x+3)(x)sin(130)=(x2+3x)=x2+3x−sin(130)12​​​​


Solve the quadratic:

−(3)±(3)2−4(1)(−12sin⁡(130))2(1)=−3±9+48sin⁡(130)2\dfrac{-(3)\pm\sqrt {(3)^2 - 4(1)(-\dfrac{12}{\sin(130)})}}{2(1)} = \dfrac{-3\pm\sqrt {9 +\dfrac{48}{\sin(130)}}}{2}2(1)−(3)±(3)2−4(1)(−sin(130)12​)​​=2−3±9+sin(130)48​​​​​


​x=2.73259... ,−5.73259...x= 2.73259... \ , -5.73259...x=2.73259... ,−5.73259...​​


A side length must be a positive value:

x=2.73 cm (2 d.p.)‾\underline{x= 2.73 \ cm \space (2\space d.p.)}x=2.73 cm (2 d.p.)​​​



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FAQs - Frequently Asked Questions

When can you use the trigonometric formula for the area of a triangle?

You can use the trigonometric formula for the area of a triangle when you have two sides and an angle between them.

Why do you use the trigonometric formula for the area of a triangle?

The trigonometric area of a triangle is used to find the area of a triangle without being given the perpendicular height.

What is the trigonometric formula for the area of a triangle?

The trigonometric formula for the area of a triangle is: 1/2(a)(b)Sin(C).

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