Trigonometric area formulas

2020-11-27T15:22:34+02:00

Trigonometric functions can also be used to calculate the areas of two-dimensional patterns if the patterns can be divided into parts consisting of right triangles.

Example 1

Derive a formula to calculate the area of a triangle from the lengths of two sides of the triangle [[$ a $]] and [[$ b $]] and the magnitude of the angle between them [[$ \alpha $]].

The height of a triangle is determined by sine:

[[$ \begin{align*} \sin \alpha &= \displaystyle\frac {h} {b} \;\;\;\;\;\;\;\; {\color {blue} {| \cdot b}} \\ b\sin \alpha &= \displaystyle\frac {bh} {b} \;\;\;\;\;\; {\color {red} {\text {Move}} \;\color {red} {h}\; \color {red} {\text {to the left of the equation and the others to the right}}}\\ -h &= -b\sin \alpha \; {\color {blue} {| \cdot \left( -1\right)}} \\ h &= b\sin \alpha \\ \end{align*} $]]

When [[$ h $]] is placed in a normal triangle area equation, a new area equation is formed. In this area equation, the height of the triangle does not have to be solved separately.

The area of a triangle

The area of a triangle is [[$ A = \displaystyle\frac {1} {2} ah = \displaystyle\frac {1} {2} ab \sin \alpha $]].

Note! The formula is applicable even when the angle [[$ \ alpha $]] is obtuse.

The corresponding surface area trigonometric formulas can also be derived for a parallelogram and trapezoid.

The area of a parallelogram

The area of a parallelogram is [[$ A = ah = ab \sin \alpha $]].

The area of a trapezoid

Area of a trapezoid is [[$ A = \displaystyle\frac {1} {2} \left(a + b \right) h = \displaystyle\frac {1} {2} \left(a + b \right) s \sin \alpha $]].

Example 2

Calculate the area of the adjacent triangle.

[[$ \begin{align*} A &= \displaystyle\frac {1} {2} ab \sin \alpha \\ \\ &= \displaystyle\frac {1} {2} \cdot 8,\!0 \cdot 7,\!0 \cdot \sin 35° \\ \\ &≈ 16 \; \text {cm}^2 \end{align*} $]]

Answer: The area of the triangle is [[$ 16 $]] cm[[$ ^2 $]].

Example 3

The lengths of a triangle's sides are [[$ 6,\!2 $]] cm and [[$ 8,\!5 $]] cm. How large is the angle between the sides when the area of the triangle is [[$ 24,\!8 $]] cm[[$ ^2 $]]?

Solution:

First, determine the sine of the angle between the sides:

[[$ \begin{align*} \displaystyle\frac {1} {2} ab \sin \alpha &= A \;\;\;\;\;\;\;\; {\color {blue} {| \cdot 2}} \\ \\ ab \sin \alpha &= 2A \;\;\;\;\;\;\ {\color {blue} {| :ab}} \\ \\ \sin \alpha &= \displaystyle\frac {2A} {ab} \\ \\ \sin \alpha &= \displaystyle\frac {2 \cdot 24,\!8 \;\text {cm}^2} {6,\!2 \;\text{cm} \cdot 8,\!5 \;\text {cm}} \\ \\ \sin \alpha &≈ 0,9412 \\ \\ \alpha &≈ 70,\!3° \\ \end{align*} $]]

The calculator gives the acute angle that implements the equation. However, there is another solution to the equation, which is [[$ 180° - 70,\!3° = 109,\!7° $]].

The calculator can still check that [[$ 180° - 70,\!3° = 109,\!7° $]].

Answer: The angle is [[$ 70,\!3° $]] or [[$ 109,\!7° $]].

In general, the following rule applies:

[[$$ \sin \alpha = \sin \left( 180° - \sin \alpha \right) $$]]