Topics
8. Right Triangles & Trigonometry
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United States of AmericaFL
Grade 912

Honors: 8.09 Areas of triangles

Lesson
Practice
Lesson

Concept summary

When we have two side lengths and the measure of the included angle, we can calculate the area of a 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} = \dfrac{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}

To prove the equation above is valid, we can start by drawing the altitude from vertex B to side AC.

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.

Using this diagram, we can:

  • Find h in terms of \angle A and side c using the sine ratio.
  • Use our existing formula for the area of a triangle A=\dfrac{1}{2} bh, to develop the formula using the 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.

Worked examples

Example 1

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

Approach

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)

Solution

\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.

Reflection

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 2

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.

Approach

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)

Solution

\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

Outcomes

MA.912.T.1.2

Solve mathematical and real-world problems involving right triangles using trigonometric ratios and the Pythagorean Theorem.

MA.912.T.1.4

Solve mathematical problems involving finding the area of a triangle given two sides and the included angle.

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