Chapter 3: Laws of Sines and Cosines
3.2 The Law of Sines
Algebra Refresher
Algebra Refresher
Convert to a decimal fraction.
- 5 inches [latex]=\underline \qquad[/latex]foot
- 10 ounces [latex]=\underline \qquad[/latex] pound
- 24 minutes [latex]=\underline \qquad[/latex]hour
- 35 seconds [latex]=\underline \qquad[/latex]minute
- 16 minutes [latex]=\underline \qquad[/latex] °
- [latex]4^{\prime} =\underline \qquad[/latex]°
- [latex]2^{\prime\prime} =\underline \qquad[/latex]°
- [latex]1^{\prime} ~5^{\prime\prime}=\underline \qquad[/latex]°
Algebra Refresher Answers
- [latex]0.41\overline{6}[/latex]
- [latex]0.625[/latex]
- [latex]0.4[/latex]
- [latex]0.58\overline{3}[/latex]
- [latex]0.2\overline{6}[/latex]
- [latex]0.0\overline{6}[/latex]
- [latex]0.000\overline{5}[/latex]
- [latex]0.0180\overline{5}[/latex]
Learning Objectives
- Use the Law of sines to find a side #1-6
- Use the Law of sines to find an angle #7-12
- Use the Law of sines to solve an oblique triangle #13-18
- Solve problems using the Law of sines #19-28
- Compute distances using parallax #29-32
- Solve problems involving the ambiguous case #33-46
Law of Sines
We have learned to use the trigonometric ratios to solve right triangles. But the trig ratios are only valid for the sides of right triangles. Can we find unknown sides or angles in an oblique triangle?
In this section and the next, we find relationships among the sides and angles of oblique triangles. These relationships are called the law of sines and the law of cosines. To derive these new rules, we use what we already know about right triangles.
Note 3.21.
Reducing a new problem to an earlier one is a frequently used technique in mathematics.
Consider the oblique triangle below. By drawing in the altitude [latex]h[/latex] of the triangle, we create two right triangles, [latex]\triangle BCD[/latex] and [latex]\triangle ABD{,}[/latex] as shown in the figure. Now we can write expressions in terms of [latex]h[/latex] for [latex]\sin A[/latex] and for [latex]\sin C{.}[/latex]
Looking at [latex]\triangle BCD{,}[/latex] we see that [latex]~~~\dfrac{h}{a} = \sin C.[/latex]
Looking at [latex]\triangle ABD{,}[/latex] we see that [latex]~~\dfrac{h}{c} = \sin A.[/latex]
Now we solve each of these equations for [latex]h{:}[/latex]
[latex]\dfrac{h}{a} = \sin C[/latex] and [latex]\dfrac{h}{c} = \sin A[/latex] Solve each equation for [latex]h[/latex].
[latex]h = a \sin C[/latex] and [latex]h = c \sin A[/latex] Equate the expressions for [latex]h[/latex].
[latex]a \sin C = c \sin A[/latex] Divide both sides by [latex]ac[/latex].
[latex]\dfrac {\sin C}{c} = \dfrac {\sin A}{a}[/latex]
We have derived a relationship between the angles [latex]\angle A[/latex] and [latex]\angle C[/latex] and their opposite sides, [latex]a[/latex] and [latex]c{.}[/latex] If we know any three of these quantities, we can find the fourth.
In a similar way, by drawing in the altitude from the vertex [latex]\angle C{,}[/latex] we can show that
[latex]\dfrac {\sin A