Topics
8. Triangle Trigonometry
topic badge
United States of America
Grades 9-12

8.07 Area of triangles

Lesson
Practice

Introduction

We will make the connection between our prior knowledge of area of triangles and use trigonometry in this lesson. Our goal is to find unknown sides of a triangle and develop a formula for the area of a triangle when the height is unknown.

Area of triangles

Recall the formula for finding the area of a triangle, A= \dfrac{1}{2}bh, where b represents a side of the triangle, called the base, and h represents the height of the triangle perpendicular to that side. Note that the base can be any side of the triangle as long as the height is drawn perpendicular to it.

Exploration

Drag the vertices to change the size of the triangle and move the slider to construct an altitude.

Loading interactive...
  1. Is the height of \triangle ABC the same as the height of the yellow triangle? Is it the same as the height of the orange triangle? Explain why or why not.
  2. Explain why the area of \triangle ABC can be represented by the equation A=\frac{1}{2}ac \sin B.
  3. Explain why the area of \triangle ABC can be represented by the equation A= \frac{1}{2}ab \sin C.

If we know the lengths of the sides of the triangle, but not the measurement of the height, there is another way to find the area of the triangle.

A triangle with vertices A B and C with corresponding sides a b c. Angle BAC is labeled with theta degrees.

In this diagram, the included angle is \angle A, so the given sides would be b and c.

\displaystyle A_{\triangle ABC} = \frac{1}{2}bc \sin\left(A\right)
\bm{A_{\triangle ABC}}
is the area of triangle ABC
\bm{b}
is the length of one of the sides
\bm{c}
is the length of another side
\bm{A}
is the angle measure of the included angle

This may also be written more generally as:

A=\dfrac{1}{2} \cdot \text{side length} \cdot \text{side length} \cdot \text{sine of included angle}

When we are using this formula to solve for the included angle, it will only give the acute answer on the calculator, so we need to remember to include both m \angle A= \theta and m \angle A= 180 \degree -\theta.

Examples

Example 1

Use the diagram to derive the formula for the area of the triangle with the sine of the included angle.

A triangle with vertices A B and C with corresponding sides a b c. Angle BAC is labeled with theta degrees.
Worked Solution
Create a strategy

We need to use a right triangle in order to write the sine of the include angle. Draw an altitude to b from \angle B to create two right triangles.

A triangle with vertices A B and C with corresponding sides a b c. Angle BAC is labeled with theta degrees. The altitude from vertex B to side AC is labeled with h.
Apply the idea

We know that the formula for the area of a triangle is A= \dfrac{1}{2}bh, and we can find the height of the triangle in terms of the sine of \angle A:

\displaystyle \sin A\displaystyle =\displaystyle \dfrac{\text{opposite}}{\text{hypotenuse}}Definition of sine
\displaystyle \sin A\displaystyle =\displaystyle \dfrac{h}{c}Substitution
\displaystyle c \sin A\displaystyle =\displaystyle hMultiply both sides of the equation by c

Then, use the area formula of a triangle and substitute in for h:

\displaystyle A_{ABC}\displaystyle =\displaystyle \dfrac{1}{2}\text{base}\cdot \text{height}Formula for area of a triangle in words
\displaystyle A_{ABC}\displaystyle =\displaystyle \dfrac{1}{2}b\cdot c \sin{A}Substitute in using the diagram and the expression for the height
\displaystyle A_{ABC}\displaystyle =\displaystyle \dfrac{1}{2}b c \sin{A}

Example 2

Calculate the area of the given triangle, rounding your answer to two decimal places.

A triangle with side lengths 15 and 11 with included angle with measure 30 degrees
Worked Solution
Create a strategy

Since we have two sides and the included angle, we can use the formula A_{\triangle ABC} = \dfrac{1}{2}bc \sin\left(A\right)

Apply the idea
\displaystyle A_{\triangle ABC}\displaystyle =\displaystyle \dfrac{1}{2}bc \sin\left(A\right)Formula
\displaystyle =\displaystyle \dfrac{1}{2}\left(11\right)\left(15\right) \sin \left( 30\right)Substitute into the formula
\displaystyle =\displaystyle 41.25Evaluate

The area of the triangle is 41.25 units^2.

Reflect and check

Since 30 \degree is a special angle measure, we do not actually need to use a calculator. This is because \sin \left(30 \degree \right)=\dfrac{1}{2}.

Example 3

For \triangle PQR, we are given:

  • A_{\triangle PQR}= 100 \text{ yd}^2
  • PQ=12 \text{ yd}
  • QR=18 \text{ yd}

Find the measure of \angle Q, rounding your answer to two decimal places.

Worked Solution
Create a strategy

We have the information we need to fill in 3 of the 4 variables, so we can solve for the one we are missing. We want to use the formula to find the area of any triangle to solve for the measure of \angle Q:

A_{\triangle PQR} = \dfrac{1}{2}\cdot PQ \cdot QR \cdot \sin\left(Q\right)

Apply the idea

Knowing only two of the side lengths will not help us determine if this triangle is acute or obtuse. Since it could be either, we need to include the obtuse angle possibility as the calculator will only give us the acute angle.

\displaystyle A_{\triangle PQR}\displaystyle =\displaystyle \dfrac{1}{2} \cdot PQ \cdot QR \cdot \sin\left(Q\right)Formula for area of a triangle
\displaystyle 100\displaystyle =\displaystyle \dfrac{1}{2}\left(12\right)\left(18\right) \sin \left( Q\right)Substitute into the formula
\displaystyle 100\displaystyle =\displaystyle 108\sin \left( Q\right)Simplify the product
\displaystyle \dfrac{100}{108}\displaystyle =\displaystyle \sin \left( Q\right)Divide both sides by 108
\displaystyle \sin^{-1}\left(\dfrac{100}{108}\right)\displaystyle =\displaystyle QApply the inverse sine to both sides
\displaystyle Q\displaystyle =\displaystyle 67.81 \degreeEvaluate on a calculator

We must include the two possible cases:

  • m \angle Q=67.81 \degree
  • m \angle Q=112.19 \degree

Example 4

Zoriana is planning to purchase topsoil for her triangular garden. Using the side lengths of her garden, how much topsoil does Zoriana need to purchase?

A triangle with side lengths 12, 20, and 14 feet.
Worked Solution
Create a strategy

We need at least one angle measure to find the area of Zoriana's garden. Use the law of cosines to solve for one angle. Let a=12, b=14, and c=20.

\displaystyle a^2\displaystyle =\displaystyle b^2+c^2-2bc\cos{A}Law of cosines
\displaystyle 12^2\displaystyle =\displaystyle 14^2 + 20^2 - 2 \left(14 \right) \left(20 \right) \cos ASubstitution
\displaystyle 144\displaystyle =\displaystyle 596 - 560 \cos AEvaluate the exponents, addition and multiplication
\displaystyle -452\displaystyle =\displaystyle -560 \cos ASubtract 596 from both sides of the equation
\displaystyle \dfrac{452}{560}\displaystyle =\displaystyle \cos ADivide both sides of the equation by -560
\displaystyle \cos^{-1} \left ( \dfrac{452}{560} \right)\displaystyle =\displaystyle \cos^{-1} \left( \cos A \right)Apply the inverse cosine function to both sides of the equation
\displaystyle 36.2 \degree\displaystyle =\displaystyle AEvaluate the inverse cosine function

Use the formula for finding the area of a triangle to calculate the square footage of the garden.

Apply the idea

The side lengths and include angle that we can use to calculate the area of the garden are b=14, c=20, and A=36.2 \degree.

\displaystyle A_{\triangle ABC}\displaystyle =\displaystyle \dfrac{1}{2}bc \sin\left(A\right)Formula
\displaystyle =\displaystyle \dfrac{1}{2}\left(14 \right) \left(20 \right) \sin\left(36.2\right)Substitution
\displaystyle =\displaystyle 82.7 \text{ ft}^2Evaluate

Zoriana will need 82.7 \text{ ft}^2 of topsoil for the garden.

Reflect and check

Consider if a different situation were given: what if we had the lengths of two sides of the garden and the non-included angle?

We can consider this case if we were given a=12, b=14, and m \angle A = 36.2 \degree. We could use the law of sines to solve for a missing angle, then find the remaining angle using the triangle sum theorem and finally the formula for finding the area of the triangle.

\displaystyle \dfrac{\sin A}{a}\displaystyle =\displaystyle \dfrac{\sin B}{b}Law of sines
\displaystyle \dfrac{\sin 36.2}{12}\displaystyle =\displaystyle \dfrac{\sin B}{14}Substitution
\displaystyle 14 \cdot \dfrac{\sin 36.2}{12}\displaystyle =\displaystyle \sin BMultiply both sides of the equation by 14
\displaystyle \sin^{-1} \left( 14 \cdot \dfrac{\sin 36.2}{12} \right)\displaystyle =\displaystyle \sin^{-1}\left( \sin B \right)Apply the inverse sine function to both sides of the equation
\displaystyle 43.6 \degree\displaystyle =\displaystyle BEvaluate the inverse sine function

We can use the triangle sum theorem to determine that m \angle C = 180 \degree - 36.2 \degree - 43.6 \degree = 100.2 \degree. Now that we have the included angle for sides a and b, we can use the formula to calculate the square footage of the garden.

\displaystyle A_{\triangle ABC}\displaystyle =\displaystyle \dfrac{1}{2}ab \sin\left(C\right)Formula
\displaystyle =\displaystyle \dfrac{1}{2}\left(12 \right) \left(14 \right) \sin\left(100.2\right)Substitution
\displaystyle =\displaystyle 82.7 \text{ ft}^2Evaluate

We found the same solution using the law of sines that we found with the law of cosines.

Idea summary

If we know the lengths of the sides of the triangle, but not the measurement of the height, there is another way to find the area of the triangle:

\displaystyle A_{\triangle ABC} = \frac{1}{2}bc \sin\left(A\right)
\bm{A_{\triangle ABC}}
is the area of triangle ABC
\bm{b}
is the length of one of the sides
\bm{c}
is the length of another side
\bm{A}
is the angle measure of the included angle

Outcomes

G.SRT.D.9 (+)

Derive the formula a = 1/2 ab sin(c) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side.

What is Mathspace

About Mathspace