Law of Sines Lesson

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)

In the case where we know the three angles of a triangle, we won't be able to find unique side lengths. This is because (AAA) will assure us of similarity, but not congruence. As an example, there are infinitely many triangles ABC of different sizes with A = 30°, B = 70°, and C = 80°

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.

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