Lesson Explainer: Finding the Area of a Triangle Using Trigonometry | Nagwa Lesson Explainer: Finding the Area of a Triangle Using Trigonometry | Nagwa

Lesson Explainer: Finding the Area of a Triangle Using Trigonometry Mathematics • First Year of Secondary School

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In this explainer, we will learn how to find the area of a triangle using the lengths of two sides and the sine of the included angle.

We have known from early on in our mathematics journey how to find the area of a triangle using its base and perpendicular height. However, we are somewhat limited by this method as these two measurements may not always be given. In this explainer, we will extend our knowledge by introducing the trigonometric formula for the area of triangles, which we now derive.

Consider triangle 𝐴𝐵𝐶 in which we know the lengths of two sides 𝑎 and 𝑏 and the measure of the angle between them, angle 𝐶. We refer to this as the included or enclosed angle. The known information is represented in bold on the figure below.

In order to apply the usual formula for the area of a triangle, we need to know the lengths of its base and perpendicular height. We draw in a line from vertex 𝐵 that is perpendicular to the base 𝐴𝐶, which we will label as .

The area of triangle 𝐴𝐵𝐶, using the usual formula, is area=𝑏2. We need to consider how to express the perpendicular height in terms of the known side lengths and the known angle. Consider triangle 𝐵𝐶𝐷 in the figure above. As this is a right triangle, we can apply the sine ratio to express in terms of 𝑎 and 𝐶. Recalling that the sine ratio is the length of the opposite side divided by the length of the hypotenuse, we have sinopphypsin𝐶=𝐶=𝑎.

Rearranging by multiplying by 𝑎 gives =𝑎𝐶.sin

We have now expressed the perpendicular height of the triangle in terms of side 𝑎 and angle 𝐶, both of which we assumed to be known. We can now substitute this expression for into the usual formula for the area of a triangle to give the trigonometric formula areasinsin=𝑏2=𝑏×𝑎𝐶2=12𝑎𝑏𝐶.

Definition: The Trigonometric Formula for the Area of Triangles

The trigonometric formula for the area of triangles is Areasin=12𝑎𝑏𝐶, where 𝑎 and 𝑏 are the lengths of two sides and 𝐶 is the measure of the included angle.

This formula is valid in both degrees and radians and can be applied to any triangle. It would be possible to work through the process of explicitly calculating the perpendicular height using trigonometry for each triangle and then applying the more basic formula, but the trigonometric formula combines these steps for us and is therefore more efficient.

It is important to note that this formula can be applied whenever we know any two sides of a triangle and their included angle. For the triangle 𝐴𝐵𝐶 above, we could equivalently express the formula as Areasin=12𝑏𝑐𝐴 or as Areasin=12𝑎𝑐𝐵.

However, it is better not to be overly concerned about the exact letters used and instead to understand what they represent in terms of the relative positioning of the sides and angle.

Let us now demonstrate how to apply this formula to calculate the area of a triangle given the lengths of two sides and the measure of the included angle.

Example 1: Using the Trigonometric Formula for the Area of Triangles

𝐴𝐵𝐶 is a triangle, where 𝐵𝐶=15cm, 𝐴𝐶=25cm, and 𝑚𝐶=41. Find the area of 𝐴𝐵𝐶, giving your answer to three decimal places.

Answer

It is helpful to produce a sketch of triangle 𝐴𝐵𝐶 as shown below (not to scale).

From our sketch, it is clear that the information we have been given consists of the lengths of two of the triangle’s sides and the measure of their included angle. We recall the trigonometric formula for the area of a triangle: areasin=12𝑎𝑏𝐶.

Substituting the side lengths of 15 cm and 25 cm and the included angle of 41 and evaluating gives areasinsin=12×15×25×41=375412=123.01106123.011.

The area of triangle 𝐴𝐵𝐶, to three decimal places, is 123.011 cm2.

In the previous example, we were explicitly given two side lengths and their included angle. In other problems, we may be given a slightly different set of information. It may then be necessary to calculate the required lengths and angles using other geometric properties, such as the angle sum in a triangle. We will now consider an example of this.

Example 2: Using the Trigonometric Formula for Areas of Triangles to Find the Area of an Isosceles Triangle

An isosceles triangle has two sides of length 48 cm and base angles of 75. Find the area of the triangle, giving the answer to three decimal places if necessary.

Answer

We begin by sketching the triangle (not to scale). We recall that the base angles of an isosceles triangle are the angles formed by each of the equal sides with the third side.

We now recall the trigonometric formula for the area of a triangle: areasin=12𝑎𝑏𝐶.

We know the lengths of two of the triangle’s sides and we can calculate the measure of their included angle, angle 𝐶 on our diagram, using the sum of angles in a triangle. By subtracting the measures of the other two angles from 180, we obtain 𝑚𝐶=1807575=30.

Substituting the two side lengths of 48 cm and the included angle of 30 into the trigonometric formula for the area of a triangle and evaluating gives areasin=12×48×48×30=12×48×48×12=576.

As this is an integer value, there is no need to round our answer to three decimal places.

The area of the triangle is 576 cm2.

In the previous problem, the included angle was one of the special angles for which the values of the three trigonometric ratios can be expressed exactly in terms of quotients and radicals. The use of such angles enables us to answer problems like these when we do not have access to a calculator.

We now summarize the key steps to follow when applying the trigonometric formula for the area of triangles.

How To: Calculating the Area of Triangles Using the Trigonometric Formula

  1. Identify a pair of side lengths and the included angle.
  2. It may be necessary to calculate any of these values using other information given in the question, such as using the sum of angles in a triangle or angles on a straight line.
  3. Substitute the values into the formula areasin=12𝑎𝑏𝐶, where 𝑎 and 𝑏 represent the lengths of the sides and 𝐶 represents the included angle.

We can also work backward when given the area of a triangle, one side length, and the measure of one angle to determine the length of the second side which encloses the angle. This will require us to form and solve an equation, as we will demonstrate in our next example.

Example 3: Finding the Length of a Side of a Triangle given Its Area, the Length of a Side, and the Measure of an Angle

𝐴𝐵𝐶 is a triangle where 𝐴𝐵=18cm, 𝑚𝐵=60, and the area of the triangle is 743 cm2. Find the length of 𝐵𝐶 giving the answer to two decimal places.

Answer

We begin by sketching triangle 𝐴𝐵𝐶 using the information given in the question.

Next, we recall the trigonometric formula for the area of a triangle: areasin=12𝑎𝑏𝐶.

We recall that 𝑎 and 𝑏 represent the lengths of any two sides and 𝐶 represents the included angle, so for our triangle we can express the area using sides 𝐴𝐵 and 𝐵𝐶 and the included angle of 60 as areasin=12×𝐴𝐵×𝐵𝐶×60.

By substituting 743 for the area and 18 for 𝐴𝐵, we can form an equation with only one unknown: 743=12×18×𝐵𝐶×60.sin

We recall that sin60=32 and solve our equation for 𝐵𝐶 by first canceling a factor of 3 from each side and then isolating 𝐵𝐶: 9×𝐵𝐶×32=7439𝐵𝐶2=74𝐵𝐶=74×29=16.44416.44.

The length of 𝐵𝐶 to two decimal places is 16.44 cm.

The problems we have seen so far have each been related to a single triangle. It is also possible to apply the trigonometric formula to calculate the areas of compound shapes involving triangles. We may need to use other skills, such as right triangle trigonometry, to calculate the missing lengths we need, as we will see in our next example.

Example 4: Finding the Area of a Compound Shape Using the Trigonometric Formula for the Area of Triangles

Find the area of the figure below giving the answer to three decimal places.

Answer

The compound shape in the figure consists of two triangles, triangle 𝐴𝐵𝐶 and triangle 𝐴𝐶𝐷. Let us consider triangle 𝐴𝐶𝐷 first. This is an equilateral triangle with a side length of 34 m and hence each of the interior angles are 60. We recall the trigonometric formula for the area of a triangle: areasin=12𝑎𝑏𝐶, where 𝑎 and 𝑏 represent the lengths of two sides and 𝐶 represents the included angle. In triangle 𝐴𝐶𝐷, every side length is 34 m and every angle is 60, so substituting these values into the formula gives areaoftrianglesinsin𝐴𝐶𝐷=12×34×34×60=57860.

Recalling that sin60=32, we have areaoftriangle𝐴𝐶𝐷=578×32=2893.

Next, we consider triangle 𝐴𝐵𝐶, which is a right triangle. We are given the measure of one other angle and we can deduce that the length of its hypotenuse, 𝐴𝐶, is 34 m. In order to apply the trigonometric formula for the area of a triangle, we first need to calculate the length of the second side that encloses angle 𝐶, side 𝐵𝐶.

In relation to angle 𝐶, side 𝐵𝐶 is the adjacent. Applying right triangle trigonometry, we have cosadjacenthypotenuse60==𝐵𝐶34.

Rearranging gives 𝐵𝐶=3460=34×12=17.cos

We are now able to apply the trigonometric formula for the area of triangles using sides 𝐵𝐶 and 𝐴𝐶 and the included angle 𝐶: areaoftrianglesin𝐴𝐵𝐶=12×17×34×60=28932.

The total area of the compound shape is the sum of the areas of the two triangles: totalareaareaoftriangleareaoftriangle=𝐴𝐶𝐷+𝐴𝐵𝐶=2893+28932=86732=750.8440750.844.

The area of the figure, to three decimal places, is 750.844 m2.

In the previous example, we applied the trigonometric formula to calculate the area of a right triangle using the lengths of two of its sides and their included angle, which in this case was not the right angle. It is interesting to note what happens if we apply the trigonometric formula using the right angle and the two sides that enclose it. These two sides are the base and perpendicular height of the triangle, as shown in the figure below.

Applying the trigonometric formula for the area of a triangle, we obtain areasin=12×𝑏××90.

Recalling that sin90=1, this simplifies to area=12𝑏, which is consistent with the usual formula for the area of a triangle using its base and perpendicular height. Thus, we have shown that if applied to a right triangle in this way, the trigonometric formula reduces to the area formula we are already familiar with.

We have seen how we can apply the trigonometric formula for the area of triangles to compound shapes, but it can also be applied to calculate the areas of certain other geometric shapes, such as parallelograms. If such shapes can be divided into triangles, then, provided we are given the necessary set of information, we can use this formula to find their area as the sum of the areas of the triangles they contain. Let us now consider an example in which we apply this formula to calculate the area of a parallelogram.

Example 5: Finding the Area of a Parallelogram Using the Trigonometric Formula for the Area of Triangles

𝐴𝐵𝐶𝐷 is a parallelogram, where 𝐴𝐵=41cm, 𝐵𝐶=27cm, and 𝑚𝐵=159. Find the area of 𝐴𝐵𝐶𝐷, giving the answer to the nearest square centimetre.

Answer

We begin by sketching the parallelogram, as shown below.

Usually when calculating the area of a parallelogram, we apply the formula areabaseperpendicularheight=×.

However, we are not given the perpendicular height of this parallelogram. Instead, we recognize that as 𝐴𝐵𝐶𝐷 is a parallelogram, each of its diagonals divide it up into two congruent triangles. Let us add the diagonal 𝐴𝐶 to our sketch.

As triangles 𝐴𝐵𝐶 and 𝐴𝐷𝐶 are congruent, they have the same area. The area of the parallelogram can therefore be calculated as twice the area of triangle 𝐴𝐵𝐶, in which we know the lengths of two sides and the measure of their included angle. We can therefore apply the trigonometric formula for the area of a triangle: areaoftrianglesin𝐴𝐵𝐶=12×41×27×159.

The area of the parallelogram is twice this: areaofsin𝐴𝐵𝐶𝐷=2×12×41×27×159.

Simplifying and evaluating gives areaofsin𝐴𝐵𝐶𝐷=41×27×159=396.713397.

The area of 𝐴𝐵𝐶𝐷, to the nearest square centimetre, is 397 cm2.

Let us finish by recapping some key points from this explainer.

Key Points

  • The area of any triangle can be calculated using the lengths of two of its sides and the sine of their included angle.
  • The trigonometric formula for the area of triangles is areasin=12𝑎𝑏𝐶, where 𝑎 and 𝑏 are the lengths of two sides and 𝐶 is the measure of the included angle.
  • When given the area of a triangle and two pieces of information from the side lengths 𝑎 and 𝑏 and the angle 𝐶, the trigonometric formula can be used to find the missing side or angle measure.
  • The trigonometric formula can also be used to calculate the areas of other geometric shapes or compound shapes which can be divided into triangles.

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