Chapter 11: Functions

11.8 Sine and Cosine Laws

Right angle trigonometry is generally limited to triangles that contain a right angle. It is possible to use trigonometry with non-right triangles using two laws: the sine law and the cosine law.

The Law of Sines

The sine law is a ratio of sines and opposite sides. The law takes the following form:

    \[\dfrac{a}{\text{sin }A}\hspace{0.25in} =\hspace{0.25in} \dfrac{b}{\text{sin }B}\hspace{0.25in} =\hspace{0.25in} \dfrac{c}{\text{sin }C}\]

Sometimes, it is written and used as the reciprocal of the above:

    \[\dfrac{\text{sin }A}{a}\hspace{0.25in} =\hspace{0.25in} \dfrac{\text{sin }B}{b}\hspace{0.25in} =\hspace{0.25in} \dfrac{\text{sin }C}{c}\]

The law of sine is used when either 2 sides and 1 opposite angle of 1 of the sides are known, or 2 angles and 1 side of 1 of the angles

The law of sine is used when either two sides and one opposite angle of one of the sides are known, or when there are two angles and one side of one of the angles. If there are two given angles of a triangle, then all three angles are known, since A^{\circ} + B^{\circ} + C^{\circ} = 180^{\circ}.

The sine law is a very useful law with one caveat in that it is possible to sometimes have two triangles (one larger and one smaller) that generate the same result. This is termed the ambiguous case and is described later in this section.

There are also textbook errors where the data given for the triangle is impossible to create. For instance:

Example 11.8.1

Can the following triangle exist?

Triangle with 120, 30 and 30 degree angles, 6 on 2 sides and 10 on the third.

If this triangle can exist, then the ratio of sines for the angles to the opposite sides should equate.

    \[\dfrac{6}{\text{sin }30^{\circ}}\hspace{0.25in} = \hspace{0.25in} \dfrac{6}{\text{sin }30^{\circ}}\hspace{0.25in} = \hspace{0.25in} \dfrac{10}{\text{sin }120^{\circ}}\]

Reducing this yields:

    \[\dfrac{6}{0.5} \hspace{0.25in} = \hspace{0.25in}\dfrac{6}{0.5} \hspace{0.25in} = \hspace{0.25in} \dfrac{10}{0.866}\]

In checking this out, we find that 12 = 12 ≠ 11.55.

This means that this triangle cannot exist.

Example 11.8.2

Find the correct length of the side opposite 120° in the triangle shown below.

Triangle with 2-30 degree angles, 1 120 degree. 2 sides with 6 and one side with 10.

For this triangle, the ratio to solve is:

    \[\dfrac{6}{\text{sin }30^{\circ}}\hspace{0.25in}=\hspace{0.25in}\dfrac{6}{\text{sin }30^{\circ}} \hspace{0.25in}=\hspace{0.25in}\dfrac{x}{\text{sin }120^{\circ}} \]

We only need to use one portion of this, so:

    \[\dfrac{6}{\text{sin }30^{\circ}}\hspace{0.25in}=\hspace{0.25in}\dfrac{x}{\text{sin }120^{\circ}}\]

Multiplying both sides of this by sin 120°, we are left with:

    \[x=\dfrac{6\text{ sin }120^{\circ}}{\text{sin }30^{\circ}}\]

This leaves us with x = 10.29.

Example 11.8.3

Find the correct length of the side opposite 120° in the triangle shown below.

Traingle with 44, 86 and 110 degree sides.

For this triangle, the ratio to solve is:

    \[\dfrac{a}{\text{sin }44^{\circ}}\hspace{0.25in}=\hspace{0.25in}\dfrac{110}{\text{sin }86^{\circ}}\]

Multiplying both sides by sin 44° leaves us with:

    \[a=\dfrac{110\text{ sin }44^{\circ}}{\text{sin }86^{\circ}}\]

Example 11.8.4

Find the unknown angle shown in the triangle shown below.

Traingle with 52 degree, 16 and 14 sides

For this triangle, the ratio to solve is:

    \[\dfrac{14}{\text{sin }A}\hspace{0.25in}=\hspace{0.25in}\dfrac{16}{\text{sin }52^{\circ}}\]

Isolating sin A yields:

    \[\text{sin }A=\dfrac{14\text{ sin }52^{\circ}}{16}\]

We now need to take the inverse sin of both sides to solve for A:

    \[\begin{array}{l} A=\text{sin}^{-1}\left(\dfrac{14\text{ sin }52^{\circ}}{16}\right) \\ \\ A=43.6^{\circ} \end{array}\]

The Ambiguous Case

It is possible, when given the right data, to create two different triangles.

1 triangle with 2 triangles inside.

You can see from the triangle shown above that it is possible to have two angles, B1 and B2, for side b. Using the sine law, you will always end up solving for B1, the angle for the largest triangle. If you are trying to solve for the smaller triangle, then you only need to subtract B1 from 180°.

For example, if B1 = 50°, then B2 = 180° − B1. This means B2 = 180° − 50° or 130°.

If the angle you solve for when using the sine law is smaller than it should be, then correct for it as we just did above.

The Law of Cosines

The Law of Cosines is the generalized law of the Pythagorean Theorem (a^2 + b^2 = c^2).

The Law of Cosines is generally written in three different forms, which are as follows:

    \[\begin{array}{l} a^2 = b^2 + c^2 - 2bc\text{ cos }A \\ b^2 = a^2 + c^2 - 2ac\text{ cos }B \\ c^2 = a^2 + b^2 - 2ab\text{ cos }C \end{array}\]

All forms revert back to one of the three regular forms of the Pythagorean theorem (a^2 = b^2 + c^2, b^2 = a^2 + c^2, c^2 = a^2 + b^2) if A, B or C is 90°, since \text{cos }90^{\circ} = 0. The following examples illustrate the usage of the cosine law in trigonometry

Example 11.8.5

Find the unknown angle shown in the triangle shown below.

Triangle with 100, 90 and 50 sides.

For this triangle:

    \[\begin{array}{l} A=\text{find} \\ \\ a=90 \\ \\ b=50 \\ \\ c=100 \end{array}\]

Note that b and c could be switched around. Now, to solve:

 

 

 

\begin{array}{rrlllll} a^2 &= &\phantom{-}b^2 &+ &c^2 &-& 2bc\text{ cos }A \\ \\ 90^2 &= &\phantom{-}50^2 &+ &100^2 &- &2(50)(100)\text{ cos } A \\ 8100& =&\phantom{-}2500 &+ &10000& - &10000\text{ cos }A \\ -2500&&-2500&-&10000&& \\ \midrule -4400& =&-10000&\text{ cos }&A&& \\ \\ \text{ cos }A& =&\dfrac{-4400}{-10000}&&&& \\ \\ B& =&\text{cos}^{-1}0.44&&&& \\ \\ B&=&63.9^{\circ}&&&& \end{array}

Example 11.8.6

Find the unknown side shown in the triangle shown below.

Triangle with sides of 96 and 60, and an angle of 70 degreesFor this triangle:

    \[\begin{array}{l} B=70^{\circ} \\ \\ a=95 \\ \\ b=\text{find} \\ \\ c=60 \end{array}\]

 

 

 

 

Note that b and c could be switched around. Now, to solve:

    \[\begin{array}{l} b^2=a^2+c^2-2ac\text{ cos }B \\ \\ b^2=95^2+60^2-2(95)(60)\text{ cos }70^{\circ} \\ b^2=9025+3600-11400(0.34202) \\ b^2=12625 - 3899 \\ b^2=8726 \\ \\ b=\sqrt{8726} \\ b=93.4 \end{array}\]

Unlike with the law of sines, there should be no ambiguous cases with the law of cosines.

Questions

Solve all unknowns in the following non-right triangles using either the law of sines or cosines.

Traingle with sides 10 and 20, angle between is 40 degrees.

Triangle with sides of 28, 28 and 20

Triangle with sides 200, 140 and 130

Triangle with sides 125, 120 and angle between of 32 degrees

Traingle with sides 18, 20 and 3

Traingle with 65 degrees, 35 degrees and side of 40

Triangle with angles of 25 and 28 degrees, side of 12

Triangle with angles of 15 and 25 degrees, side of 10

Traingle with angles of 10 and 70 degrees, and a side of 8 cm

Triangle with sides of 28 and 20 cm, angle between of 130 degrees

Triangle with sides of 30, 20 and 15 m

Triangle with angles of 95 and 20 degrees, side of 8 m

Triangle with sides of 10, 16 and 8 cm

Triangle with sides of 20 and 24 cm, 15 degree angle

Triangle with sides of 22, 20 and 10 m

Triangle with angles of 28 and 25 degrees, 20 m side

 

<a class=”internal” href=”/intermediatealgebraberg/back-matter/answer-key-11-8/”>Answer Key 11.8

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