Lesson Objectives

- Learn how to apply the law of sines (SAA)
- Learn how to apply the law of sines (ASA)
- Learn how to find the area of a triangle (SAS)

## Oblique Triangles and the Law of Sines

### Solving Oblique Triangles

A triangle that is not a right triangle is known as an oblique triangle. The measures of the three sides and three angles of a triangle can be found when at least one side is known and any two other measures are known. This leads to four possible scenarios:- Case 1: One side and two angles are known (SAA) or (ASA)
- Case 2: Two sides and one angle not included between the two sides are known (SSA)
- This scenario may lead to more than one triangle

- Case 3: Two sides and the angle included between the two sides are known (SAS)
- Case 4: Three sides are known (SSS)

### Law of Sines

To think about where the law of sines comes from, let's start with an oblique triangle, such as the acute triangle shown below: In triangle ADB: $$\text{sin}\hspace{.1em}\text{A}=\frac{h}{c}$$ In triangle BDC: $$\text{sin}\hspace{.1em}\text{C}=\frac{h}{a}$$ If we solve for h in each case we get: $$h=\text{sin}\hspace{.1em}A \cdot c$$ $$h=\text{sin}\hspace{.1em}C \cdot a$$ Since we have h in each case, we can set the two expressions equal to each other. $$\text{sin}\hspace{.1em}A \cdot c=\text{sin}\hspace{.1em}C \cdot a$$ Let's now divide each side by: sin A sin C: $$\frac{\text{sin}\hspace{.1em}A \cdot c}{\text{sin}\hspace{.1em}A \hspace{.1em}\cdot \text{sin}\hspace{.1em}C}=\frac{\text{sin}\hspace{.1em}C \cdot a}{\text{sin}\hspace{.1em}A \hspace{.1em}\cdot \text{sin}\hspace{.1em}C}$$ $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{c}{\text{sin}\hspace{.1em}C}$$ Using a similar thought process, we could also show: $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}$$ $$\frac{b}{\text{sin}\hspace{.1em}B}=\frac{c}{\text{sin}\hspace{.1em}C}$$ We can write this in a more compact form. In any triangle ABC, with sides a, b, and c: $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}=\frac{c}{\text{sin}\hspace{.1em}C}$$ To summarize, the law of sines tells us that the lengths of the sides in a triangle are proportional to the sines of the measures of the angles opposite them.### Case 1: Solving ASA or SAA Triangles

When two angles and one side are known, then we can simply use the law of sines to solve the triangle. Let's look at a few examples. Note, in our examples, we will use triangles named ABC, where A, B, and C are angles. The sides opposite of each angle are named with corresponding lowercase letters. In our case, we will see a, b, and c as the sides.Example #1: Solve triangle ABC. Note: Uppercase letters denote angles, whereas, lowercase letters denote sides. $$B=46°, C=69°, a=34 \hspace{.1em}\text{cm}$$ To start this problem, it's best to draw a sketch. The measure of angle A can easily be found by the angle sum property of a triangle. This tells us that the sum of the angles in a triangle is 180°. $$A=180° - 46° - 69°=65°$$ Additionally, we need to find sides b and c. Let's start with side b. For this, we will use our law of sines. $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}=\frac{c}{\text{sin}\hspace{.1em}C}$$ Since we know the measures of angles A and B, and the length of side a, we can use the following to obtain b. $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}$$ $$\frac{34 \hspace{.1em}\text{cm}}{\text{sin}\hspace{.1em}65°}=\frac{b}{\text{sin}\hspace{.1em}46°}$$ $$b=\frac{34 \hspace{.1em}\text{cm}\cdot \text{sin}\hspace{.1em}46°}{\text{sin}\hspace{.1em}65°}$$ $$b ≈ 27 \hspace{.1em}\text{cm}$$ Here, we rounded to the nearest tenth to approximate our answer.

Let's now find side c using a similar process. $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{c}{\text{sin}\hspace{.1em}C}$$ $$\frac{34 \hspace{.1em}\text{cm}}{\text{sin}\hspace{.1em}65°}=\frac{c}{\text{sin}\hspace{.1em}69°}$$ $$c=\frac{34 \hspace{.1em}\text{cm}\cdot \text{sin}\hspace{.1em}69°}{\text{sin}\hspace{.1em}65°}$$ $$c ≈ 35 \hspace{.1em}\text{cm}$$ Here, we rounded to the nearest tenth to approximate our answer.

Example #2: Solve triangle ABC. Note: Uppercase letters denote angles, whereas, lowercase letters denote sides. $$A=52°, B=67°, a=18 \hspace{.1em}\text{cm}$$ To start this problem, it's best to draw a sketch. The measure of angle C can easily be found by the angle sum property of a triangle. This tells us that the sum of the angles in a triangle is 180°. $$C=180° - 52° - 67°=61°$$ Additionally, we need to find sides b and c. Let's start with side b. For this, we will use our law of sines. $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}=\frac{c}{\text{sin}\hspace{.1em}C}$$ Since we know the measures of angles A and B, and the length of side a, we can use the following to obtain b. $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}$$ $$\frac{18 \hspace{.1em}\text{cm}}{\text{sin}\hspace{.1em}52°}=\frac{b}{\text{sin}\hspace{.1em}67°}$$ $$b=\frac{18 \hspace{.1em}\text{cm}\cdot \text{sin}\hspace{.1em}67°}{\text{sin}\hspace{.1em}52°}$$ $$b ≈ 21 \hspace{.1em}\text{cm}$$ Here, we rounded to the nearest tenth to approximate our answer.

Let's now find side c using a similar process. $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{c}{\text{sin}\hspace{.1em}C}$$ $$\frac{18 \hspace{.1em}\text{cm}}{\text{sin}\hspace{.1em}52°}=\frac{c}{\text{sin}\hspace{.1em}61°}$$ $$c=\frac{18 \hspace{.1em}\text{cm}\cdot \text{sin}\hspace{.1em}61°}{\text{sin}\hspace{.1em}52°}$$ $$c ≈ 20 \hspace{.1em}\text{cm}$$ Here, we rounded to the nearest tenth to approximate our answer.

Example #3: Solve triangle ABC. Note: Uppercase letters denote angles, whereas, lowercase letters denote sides. $$A=83°, C=53°, c=8 \hspace{.1em}\text{yd}$$ To start this problem, it's best to draw a sketch. The measure of angle B can easily be found by the angle sum property of a triangle. This tells us that the sum of the angles in a triangle is 180°. $$B=180° - 83° - 53°=44°$$ Additionally, we need to find sides a and b. Let's start with side a. For this, we will use our law of sines. $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}=\frac{c}{\text{sin}\hspace{.1em}C}$$ Since we know the measures of angles A and C, and the length of side c, we can use the following to obtain a. $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{c}{\text{sin}\hspace{.1em}C}$$ $$\frac{a}{\text{sin}\hspace{.1em}83°}=\frac{8 \hspace{.1em}\text{yd}}{\text{sin}\hspace{.1em}53°}$$ $$a=\frac{8 \hspace{.1em}\text{yd}\cdot \text{sin}\hspace{.1em}83°}{\text{sin}\hspace{.1em}53°}$$ $$a ≈ 9.9 \hspace{.1em}\text{yd}$$ Here, we rounded to the nearest tenth to approximate our answer.

Let's now find side b using a similar process. $$\frac{b}{\text{sin}\hspace{.1em}B}=\frac{c}{\text{sin}\hspace{.1em}C}$$ $$\frac{b}{\text{sin}\hspace{.1em}44°}=\frac{8 \hspace{.1em}\text{yd}}{\text{sin}\hspace{.1em}53°}$$ $$b=\frac{8 \hspace{.1em}\text{yd}\cdot \text{sin}\hspace{.1em}44°}{\text{sin}\hspace{.1em}53°}$$ $$b ≈ 7 \hspace{.1em}\text{yd}$$ Here, we rounded to the nearest tenth to approximate our answer.

## Area of a Triangle (SAS)

The method used to derive the law of sines can also be used to derive a formula to find the area of a triangle. $$\text{Area}=\frac{1}{2}bh$$ Where b is the base, and h is the height.In some cases, the height will not be known. Let's revisit our acute triangle from earlier in the lesson: Let's consider our right triangle ADB: $$\text{sin}\hspace{.1em}A=\frac{h}{c}$$ Solve for h: $$h=\text{sin}\hspace{.1em}A \cdot c$$ Now, let's plug in for h in our formula for the area of a triangle. $$\text{Area}=\frac{1}{2}b \cdot \text{sin}\hspace{.1em}A \cdot c$$ Similarly, we could have used any other pair of sides and the angle between them. In any triangle ABC, the area is given by the following formulas: $$\text{Area}=\frac{1}{2}bc \cdot \text{sin}\hspace{.1em}\text{A}$$ $$\text{Area}=\frac{1}{2}ab \cdot \text{sin}\hspace{.1em}\text{C}$$ $$\text{Area}=\frac{1}{2}ac \cdot \text{sin}\hspace{.1em}\text{B}$$ We can summarize this formula as the area is one-half the product of the lengths of the two sides and the sine of the angle included between them. Let's look at a few examples.

Example #4: Find the area of the triangle. $$b=5\text{km}, a=10\text{km}, C=121°$$ To start this problem, it's best to draw a sketch. Here, we have a = 10 km, b = 5 km, and C = 121°. Since we are working with sides a and b, we will use the sine of the angle included between them (sine of C). $$\text{Area}=\frac{1}{2}10 \hspace{.1em}\text{km}\cdot 5 \hspace{.1em}\text{km}\cdot \text{sin}\hspace{.1em}(121°)$$ $$\text{Area}=\frac{50}{2}\text{km}^{2}\cdot \text{sin}\hspace{.1em}(121°)$$ $$\text{Area}=25\text{km}^{2}\cdot \text{sin}\hspace{.1em}(121°)$$ $$\text{Area}≈ 21.4 \hspace{.1em}\text{km}^{2}$$ Here, we rounded to the nearest tenth to approximate our answer.

Example #5: Find the area of the triangle. $$a=11\text{ft}, C=69°, B=76°$$ To start this problem, it's best to draw a sketch. Here, we have a = 11 ft, B = 76°, and C = 69°. For this scenario, we need to find one additional side. Let's find b. $$A=180° - 69° - 76°=35°$$ $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}$$ $$\frac{11 \hspace{.1em}\text{ft}}{\text{sin}\hspace{.1em}35°}=\frac{b}{\text{sin}\hspace{.1em}76°}$$ $$b=\frac{11 \hspace{.1em}\text{ft}\cdot \text{sin}\hspace{.1em}76°}{\text{sin}\hspace{.1em}35°}$$ $$b ≈ 18.6\hspace{.1em}\text{ft}$$ Now, let's return to our problem. $$\text{Area}=\frac{1}{2}11 \text{ft}\cdot 18.6 \text{ft}\cdot \text{sin}(69°)$$ $$\text{Area}≈ 95.5\text{ft}^{2}$$ Here, we rounded to the nearest tenth to approximate our answer.

#### Skills Check:

Example #1

Find side b in triangle ABC. Round to the nearest tenth. $$A=31°, B=27°, a=34\text{ft}$$

Please choose the best answer.

A

$$b=12.9\text{ft}$$

B

$$b=30\text{ft}$$

C

$$b=31\text{ft}$$

D

$$b=28\text{ft}$$

E

$$b=27\text{ft}$$

Example #2

Find side b in triangle ABC. Round to the nearest tenth. $$A=13°, B=18°, c=55\text{km}$$

Please choose the best answer.

A

$$b=36\text{km}$$

B

$$b=33\text{km}$$

C

$$b=32\text{km}$$

D

$$b=31.8\text{km}$$

E

$$b=34.6\text{km}$$

Example #3

Find the area of triangle ABC. Round to the nearest tenth. $$C=21°, B=143°, b=16\text{mi}$$

Please choose the best answer.

A

$$17.1 \text{mi}^{2}$$

B

$$24.5 \text{mi}^{2}$$

C

$$19.5 \text{mi}^{2}$$

D

$$22.7 \text{mi}^{2}$$

E

$$20.9 \text{mi}^{2}$$

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