The Law of Sines

Section 4.1 The Law of Sines

Recall Example 2.4.18 in Section 2.4 where Alingano Maisu is sailing to her new home in Satawal. For the first leg of the voyage, we calculated that to sail from Johnston Atoll to Majuro, we would need to head towards House ʻĀina Kona (\(27.4^{\circ}\)). However, in reality, currents can push the waʻa off course. Let’s consider a scenario where a current of 1 knot is flowing in the direction of the house Komohana (West). If we were to face Majuro and sail straight, the current would push us to the west of our destination. To ensure we arrive at Majuro, our apparent heading will need to be adjusted accordingly, as illustrated below.

Now, suppose we sail at 5 knots. What heading should we aim for to compensate for a 1-knot current? We note that by alternate interior angles, \(A=27.4^{\circ}\text{.}\) Let’s represent this situation with a triangle where the sides represent speed of the waʻa and the current, allowing us to disregard the distance traveled.

It’s important to note that this triangle is not a right triangle; instead, it’s classified as an oblique triangle, meaning it does not have any right angles. Triangles can be further categorized as either obtuse or acute. An obtuse triangle contains one angle greater than \(90^{\circ}\text{,}\) while an acute triangle consists of all angles less than \(90^{\circ}\text{.}\) Solving such triangles requires different methods than those used for solving the right triangle in previous chapters. In this chapter, we’ll learn techniques for solving oblique triangles, starting with the Law of Sines and the Law of Cosines.

To solve an oblique triangle, we need to be given three pieces of information, which can take the following forms:

One side and two angles.
- Angle-Side-Angle (ASA): Two angles and the side between them are known.
- Side-Angle-Angle (SAA): Two angles and a side that is not between them are known.
Two sides and one angle.
- Side-Side-Angle (SSA): Two sides and the angle that is not between them are known.
- Side-Angle-Side (SAS): Two sides and the angle between them are known.
Three sides.
- Side-Side-Side (SSS): All three sides of the triangle are known.

In this section, we explore techniques for solving triangles when provided with one side and two angles (ASA or SAA) and when given two sides along with the angle not between them (SSA). Triangles falling under the categories of SAS and SSS will be discussed in more detail in the Section 4.2.

Subsection 4.1.1 Law of Sines

The Law of Sines states that for any triangle, the ratios of the lengths of the sides to the sine of their corresponding opposite angles are equal. When provided with certain information about the sides and angles of a triangle, this fundamental principle becomes a valuable tool for determining the remaining sides and angles.

Theorem 4.1.1. Law of Sines.

In triangle \(ABC\text{,}\) we have

\begin{equation*} \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\text{.} \end{equation*}

which may also be written in the form

\begin{equation*} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\text{.} \end{equation*}

The first form is more convenient when we are trying to find an angle and the second form is more convenient when we are trying to find a side.

Proof.

In any triangle, we can draw an altitude, a line through a vertex and perpendicular to the side opposite the vertex.

Using the right triangles, we have \(\sin A=\frac{h}{b}\) and \(\sin B=\frac{h}{a}\text{.}\) Solving both equations for \(h\text{,}\) we get \(h=b\sin A\) and \(h=a\sin B\text{.}\) Since these equations both equal \(h\text{,}\) we can set them equal to each other:

\begin{equation*} b\sin A = a\sin B\text{.} \end{equation*}

Dividing both sides by \(ab\text{,}\) we get

\begin{equation*} \frac{\sin A}{a}=\frac{\sin B}{b}\text{.} \end{equation*}

Using the same method, we can get:

\begin{equation*} \frac{\sin A}{a}=\frac{\sin C}{c}\mbox{ and } \frac{\sin B}{b}=\frac{\sin C}{c}\text{.} \end{equation*}

Putting these together, we get

\begin{equation*} \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\text{.} \end{equation*}

Remark 4.1.2.

When solving triangles, it is often useful to use the fact that the sum of the angles in a triangle equals \(180^{\circ}\text{.}\) That is,

\begin{equation*} A+B+C=180^{\circ}\text{.} \end{equation*}

Subsection 4.1.2 Solving ASA Triangles

Example 4.1.3. Using Law of Sines (ASA).

The waʻa Kānehūnāmoku is returning from a morning sail, training students in the ways of sailing. Two observers standing 1,000 feet apart on the shore at Kāneʻohe Bay spot Kānehūnāmoku. Observer \(A\) determines that the angle between the waʻa and Observer \(B\) is \(80^{\circ}\) while Observer \(B\) determines that the angle between the waʻa and Observer \(A\) is \(60^{\circ}\text{.}\) How far is Kānehūnāmoku from each observer and what is the angle formed by the two observers (round the answer to the nearest foot)?

Solution.

In this triangle, we know the values of two angles and the side between them, so we have an Angle-Side-Angle (ASA) triangle. We know that \(A+B+C=180^{\circ}\text{.}\) So \(80^{\circ}+60^{\circ}+C=140^{\circ}+C=180^{\circ}\) so \(C=40^{\circ}\text{.}\) To solve for \(a\text{,}\) we will use the Law of Sines. Either form for the Law of Sines can be used to solve this problem, but since we are solving for a side, it would be easier and require fewer steps if we use the second form for the Law of Sines.

\begin{align*} \frac{a}{\sin A} =\amp \frac{c}{\sin C}\\ \frac{a}{\sin 80^{\circ}} =\amp \frac{1,000}{\sin 40^{\circ}} \end{align*}

Multiplying both sides by \(\sin 80^{\circ}\text{,}\)

\begin{equation*} a=\frac{1,000\sin 80^{\circ}}{\sin 40^{\circ}}\approx1,532\mbox{ ft}\text{.} \end{equation*}

To solve for \(b\text{,}\) we have two options to use for the Law of Sines:

\begin{equation*} \frac{b}{\sin B}=\frac{a}{\sin A}\qquad\mbox{or} \qquad\frac{b}{\sin B}=\frac{c}{\sin C}\text{.} \end{equation*}

Because our answer for \(a\) was rounded, using the first option may lead to rounding errors so it is best to use the second option because it has known values that we were originally given in the problem (\(c=1,000\)). Thus we will use

\begin{equation*} \frac{b}{\sin B}=\frac{c}{\sin C}\text{.} \end{equation*}

So we get

\begin{equation*} b=\frac{c\sin B}{\sin C}=\frac{1,000\sin 60^{\circ}}{\sin 40^{\circ}}\approx1,347\mbox{ ft}\text{.} \end{equation*}

Thus Kānehūnāmoku is 1,347 feet from Observer A and 1,532 feet from Observer B, and the angle between the two observers is \(40^{\circ}\text{.}\)

Subsection 4.1.3 Solving SAA Triangles

Example 4.1.4. Using Law of Sines (SAA).

For the triangle shown below with \(A=120^{\circ}\text{,}\) \(C=20^{\circ}\text{,}\) and \(a=5\text{,}\) find the remaining sides. Round the answer to two decimal places.

Solution.

In this triangle, we know the values of two angles and the side between them, so we have a Side-Angle-Angle (SAA) triangle. The third angle can be calculated by subtracting the known angles from \(180^{\circ}\text{:}\)

\begin{equation*} B=180^{\circ}-A-C=180^{\circ}-120^{\circ}-20^{\circ}=40^{\circ}\text{.} \end{equation*}

By using the first form of the Law of Sines

\begin{equation*} \frac{b}{\sin B}=\frac{a}{\sin A}\qquad\mbox{and} \qquad\frac{c}{\sin C}=\frac{a}{\sin A} \end{equation*}

we have

\begin{equation*} b=\frac{a\sin B}{\sin A}=\frac{5\sin40^{\circ}}{\sin120^{\circ}}\approx3.71 \end{equation*}

and

\begin{equation*} c=\frac{a\sin C}{\sin A}=\frac{5\sin20^{\circ}}{\sin120^{\circ}}\approx1.97\text{.} \end{equation*}

Subsection 4.1.4 Solving SSA Triangles: The Ambiguous Case

Although the Law of Sines can be used to solve oblique triangles, there may be cases that would give us one triangle, two triangles, or even no solution. Such cases are known as the ambiguous case. This occurs when we know two sides and an angle that is opposite to one of the given sides.

Definition 4.1.5. The Ambiguous Case.

Consider a triangle where \(a\text{,}\) \(b\text{,}\) and \(A\) are given. The altitude is \(h=b\sin A\text{.}\) We have the following possibilities:

\(A\) is acute and \(a\lt h=b\sin A\text{:}\) No triangle
\(A\) is acute and \(a=h=b\sin A\text{:}\) One right triangle
\(A\) is acute and \(h=b\sin A\lt a\lt b\text{:}\) Two triangles (\(\triangle AB_1C_1\) and \(\triangle AB_2C_2\))
\(A\) is acute and \(a\geq b\text{:}\) One triangle
\(A\) is obtuse and \(a\leq b\text{:}\) No triangle
\(A\) is obtuse and \(a>b\text{:}\) One triangle

Remark 4.1.6. Generalizing the Ambiguous Case.

Not every time we encounter a SSA triangle will we be given the angle \(A\) and sides \(a\) and \(b\text{.}\) Sometimes we may be given angle \(B\) and sides \(a\) and \(b\text{.}\) In these cases, we cannot use the equations and inequalities listed above to identify how many triangles we will get.

To generalize the ambiguous case of SSA, we are given a triangle with:

a known angle, \(\theta\text{;}\)
a side adjacent to \(\theta\text{;}\)
a side opposite to \(\theta\text{.}\)

If we calculate the altitude \(h=\mbox{(adjacent side)}\cdot\sin\theta\text{,}\) then Table 4.1.7 summarizes the possible cases.

Table 4.1.7. Possibilities for the ambiguous case (SSA)

Case	\(\theta\)	Condition on Opposite Side	Number of Triangles
1	acute	opposite side \(\lt\) altitude	None
2	acute	opposite side = altitude	One (right triangle)
3	acute	altitude \(\lt\) opposite side \(\lt\) adjacent side	Two
4	acute	opposite side \(\geq\) adjacent side	One
5	obtuse	opposite side \(\leq\) adjacent side	None
6	obtuse	opposite side \(>\) adjacent side	One

Example 4.1.8. Using Law of Sines (SSA) one-solution.

We can now return to the example at the start of the section where the Alingano Maisu must account for a current and we had the following triangle:

Which house do we need to sail into to account for the current?

Solution.

Referring to Table 4.1.7, our known angle (\(A=27.4^{\circ}\)) is acute, the side opposite of our angle (\(a=5\)) is greater than the side adjacent to the angle (\(c=1\)) so we have Case 4 and know that we have one triangle.

Using the first form of the Law of Sines

\begin{equation*} \frac{\sin A}{a}=\frac{\sin C}{c} \end{equation*}

we get

\begin{equation*} \sin C=c\cdot\frac{\sin A}{a}\approx1\cdot\frac{\sin 27.4^{\circ}}{5}\approx0.092\text{.} \end{equation*}

Taking the inverse sine we get

\begin{equation*} C=\sin^{-1}0.092\approx5.3^{\circ}\text{.} \end{equation*}

Now that we know the value of \(C\text{,}\) we will need to add that to the heading that we determined in the last section (\(207.4^{\circ}\)) to get \(5.3^{\circ}+207.4^{\circ}=212.7^{\circ}\text{.}\) Next we refer to the Star Compass with angles (Figure 1.2.4) to conclude we will need to sail towards the House Noio Kona.

Example 4.1.9. Using Law of Sines (SSA) no solution.

Solve the triangle if \(C=70^{\circ}\text{,}\) \(b=6\text{,}\) and \(c=5\text{.}\)

Solution.

The altitude of the triangle is \(h=b\sin C=6\sin70^{\circ}\approx5.64\text{.}\) Since \(C\) is an acute angle and the altitude (\(h\approx5.64\)) is greater than the side opposite of the angle (\(c=5)\text{,}\) we have Case 1 of Table 4.1.7, thus there is no triangle.

Example 4.1.10. Using Law of Sines (SSA) two solutions.

Solve the triangle if \(a=6\text{,}\) \(b=7\text{,}\) and \(A=40^{\circ}\text{.}\)

Solution.

The altitude of the triangle is \(h=b\sin A=7\sin40^{\circ}\approx4.50\text{.}\) Since \(A\) is an acute angle and the altitude (\(h\approx4.50\)) is less than the opposite side (\(a=6\)) which is less than the adjacent side (\(b=7)\text{,}\) this satisfies Case 3 of Table 4.1.7, thus there are two triangles.

Using the first form for the Law of Sines

\begin{equation*} \frac{\sin A}{a}=\frac{\sin B}{b} \end{equation*}

we get

\begin{equation*} \sin B=\frac{b\sin A}{a}=\frac{7\sin40^{\circ}}{6}\approx0.75\text{.} \end{equation*}

Here we have that \(\sin B\) is positive and we know that sine is positive in Quadrant I and II so our value for \(B\) is between \(0^{\circ}\) and \(180^{\circ}\text{.}\) Taking the inverse sine we get our first angle

\begin{equation*} B_1=\sin^{-1}(0.75)\approx48.58^{\circ}\text{,} \end{equation*}

which is in Quadrant I. To calculate the angle in Quadrant II, we subtract the reference angle from \(180^{\circ}\) to get

\begin{equation*} B_2=180^{\circ}-B_1=180^{\circ}-48.58^{\circ}=131.42^{\circ}\text{.} \end{equation*}

So we now have two triangles: \(AB_1C_1\) and \(AB_2C_2\text{.}\)

To solve triangle \(AB_1C_1\text{,}\) be begin by finding \(C_1\text{:}\)

\begin{equation*} C_1\approx180^{\circ}-A-B_1=180^{\circ}-40^{\circ}-48.58^{\circ}=91.42^{\circ}\text{.} \end{equation*}

To find the side \(c_1\text{,}\) we use the second form for the Law of Sines

\begin{equation*} \frac{c_1}{\sin C_1}=\frac{a}{\sin A}\text{.} \end{equation*}

Thus

\begin{equation*} c_1=\frac{a\sin C_1}{\sin A}\approx\frac{6\sin48.58^{\circ}}{\sin40^{\circ}}\approx9.33\text{.} \end{equation*}

To solve triangle \(AB_2C_2\text{,}\) we begin by finding \(C_2\)

\begin{equation*} C_2\approx180^{\circ}-A-B_2=180^{\circ}-40^{\circ}-131.42^{\circ}=8.58^{\circ}\text{.} \end{equation*}

To find the side \(c_2\text{,}\) we use the second form for the Law of Sines to get

\begin{equation*} c_2=\frac{a\sin C_2}{\sin A}\approx\frac{6\sin8.58^{\circ}}{\sin40^{\circ}}\approx1.39\text{.} \end{equation*}

Subsection 4.1.5 Area of an oblique triangle

After heavy use, we determine it’s time to replace the peʻa (sail) of our waʻa. Before we buy the new peʻa, we need to determine its area. We know the length of the ʻōpeʻa (spar) is \(b=18\) feet, the length of the paepae (boom) is \(a=15\) feet, and the tack or the angle between them is \(C=44^{\circ}\text{.}\) What is the area of the peʻa?

At this point, we don’t have a formula for the area of an oblique triangle. However, we can apply the sine function to an oblique triangle to develop a new formula for the area of a triangle.

Definition 4.1.11. Area of a Triangle.

Given two sides of a triangle and their included angle, the area of the triangle is given by:

\begin{align*} \mbox{Area} \amp = \frac{1}{2}ab\sin C\\ \amp = \frac{1}{2}ac\sin B \\ \amp = \frac{1}{2}bc\sin A \text{.} \end{align*}

In other words, the area of a triangle equals one half of the product of two sides and the sine of their included angle.

Proof.

Recall that the area of any triangle is \(\mbox{Area} =\frac{1}{2}bh\) where \(b\) is the base and \(h\) is the height of a triangle. In a right triangle, the height is the length of one side. However, in an oblique triangle, the height is not immediately known.

If the angle \(C\) is acute, then using the property that

\begin{equation*} \sin C=\frac{\mbox{opposite} }{\mbox{adjacent} }=\frac{h}{a}\text{,} \end{equation*}

the height is given by

\begin{equation*} h=a\sin C\text{.} \end{equation*}

If the angle \(C\) is obtuse, we use the reference angle for \(C\text{,}\) \(180^{\circ}-C\) to get

\begin{equation*} h=a\sin(180^{\circ}-C)\text{.} \end{equation*}

By the difference formula we get \(\sin(180^{\circ}-C)=\sin C\) and thus

\begin{equation*} h=a\sin C\text{.} \end{equation*}

In a right triangle, \(C=90^{\circ}\) and since \(\sin90^{\circ}=1\text{,}\) we can also write

\begin{equation*} h=a\sin C=a\sin90^{\circ}=a\cdot1=a\text{.} \end{equation*}

Thus regardless of the triangle, we get

\begin{equation*} \mbox{Area} =\frac{1}{2}(\mbox{base} )(\mbox{height} )=\frac{1}{2}b(a\sin C)=\frac{1}{2}ab\sin C\text{.} \end{equation*}

The other formulas are obtained using a similar method.

Example 4.1.12. Area of a sail.

We can now calculate the area of the sail.

Solution.

Using the formula, the area is

\begin{equation*} \mbox{Area} =\frac{1}{2}ab\sin C=\frac{1}{2}(15)(18)\sin44^{\circ}\approx93.8\mbox{ft} ^2 \end{equation*}

Thus we will need to order a sail that is 93.8ft\(^2\text{.}\)

Exercises 4.1.6 Exercises

Exercise Group.

Use the Law of Sines to solve each triangle. Express angles in degrees rounded to one decimal and sides rounded to two decimals.

1.

\(a=12\text{,}\) \(B=45^{\circ}\text{,}\) \(C=60^{\circ}\)

Answer.

\(b\approx8.78\text{,}\) \(c\approx10.76\text{,}\) \(A=75^{\circ}\)

2.

\(a=10\text{,}\) \(A=35^{\circ}\text{,}\) \(B=60^{\circ}\)

Answer.

\(b\approx15.10\text{,}\) \(c\approx17.37\text{,}\) \(C=85^{\circ}\)

3.

\(b=12\text{,}\) \(A=45^{\circ}\text{,}\) \(B=75^{\circ}\)

Answer.

\(a\approx8.78\text{,}\) \(c\approx10.76\text{,}\) \(C=60^{\circ}\)

4.

\(c=15\text{,}\) \(A=65^{\circ}\text{,}\) \(B=45^{\circ}\)

Answer.

\(a\approx14.47\text{,}\) \(b\approx11.29\text{,}\) \(C=70^{\circ}\)

5.

\(c=14\text{,}\) \(A=50^{\circ}\text{,}\) \(C=80^{\circ}\)

Answer.

\(a\approx10.89\text{,}\) \(b\approx10.89\text{,}\) \(B=50^{\circ}\)

6.

\(b=15\text{,}\) \(A=50^{\circ}\text{,}\) \(C=70^{\circ}\)

Answer.

\(a\approx13.27\text{,}\) \(c\approx16.28\text{,}\) \(B=60^{\circ}\)

Exercise Group.

Use the Law of Sines to solve each triangle. Express angles in degrees rounded to one decimal and sides rounded to two decimals.

7.

\(a=8\text{,}\) \(A=50^{\circ}\text{,}\) \(C=80^{\circ}\)

Answer.

\(b=8\text{,}\) \(c\approx10.28\text{,}\) \(B=50^{\circ}\)

8.

\(b=12\text{,}\) \(B=60^{\circ}\text{,}\) \(C=30^{\circ}\)

Answer.

\(a\approx13.86\text{,}\) \(c\approx6.93\text{,}\) \(A=90^{\circ}\)

9.

\(c=10\text{,}\) \(A=40^{\circ}\text{,}\) \(C=70^{\circ}\)

Answer.

\(a\approx6.84\text{,}\) \(b=10\text{,}\) \(B=70^{\circ}\)

10.

\(a=15\text{,}\) \(A=40^{\circ}\text{,}\) \(B=100^{\circ}\)

Answer.

\(b\approx22.98\text{,}\) \(c=15\text{,}\) \(C=40^{\circ}\)

11.

\(b=20\text{,}\) \(B=80^{\circ}\text{,}\) \(A=20^{\circ}\)

Answer.

\(a\approx6.94\text{,}\) \(c=20\text{,}\) \(C=80^{\circ}\)

12.

\(a=8\text{,}\) \(A=40^{\circ}\text{,}\) \(B=75^{\circ}\)

Answer.

\(b\approx12.02\text{,}\) \(c\approx11.28\text{,}\) \(C=65^{\circ}\)

13.

\(b=8\text{,}\) \(A=30^{\circ}\text{,}\) \(C=50^{\circ}\)

Answer.

\(a\approx4.06\text{,}\) \(c\approx6.22\text{,}\) \(B=100^{\circ}\)

14.

\(c=15\text{,}\) \(A=70^{\circ}\text{,}\) \(B=45^{\circ}\)

Answer.

\(a\approx15.55\text{,}\) \(b\approx11.70\text{,}\) \(C=65^{\circ}\)

15.

\(a=10\text{,}\) \(B=60^{\circ}\text{,}\) \(C=70^{\circ}\)

Answer.

\(b\approx11.31\text{,}\) \(c\approx12.27\text{,}\) \(A=50^{\circ}\)

16.

\(b=14\text{,}\) \(A=45^{\circ}\text{,}\) \(C=80^{\circ}\)

Answer.

\(a\approx12.09\text{,}\) \(c\approx16.83\text{,}\) \(B=55^{\circ}\)

17.

\(c=20\text{,}\) \(A=60^{\circ}\text{,}\) \(B=75^{\circ}\)

Answer.

\(a\approx24.49\text{,}\) \(b\approx27.32\text{,}\) \(C=45^{\circ}\)

18.

\(c=18\text{,}\) \(A=70^{\circ}\text{,}\) \(B=50^{\circ}\)

Answer.

\(a\approx19.53\text{,}\) \(b\approx15.92\text{,}\) \(C=60^{\circ}\)

Exercise Group.

For each problem, determine the relevant case from Table 4.1.7 and whether it results one triangle, two triangles, or none. Then solve using the Law of Sines, with angles in degrees rounded to one decimal place and sides to two decimals.

19.

\(c=5\text{,}\) \(b=4\text{,}\) \(C=120^{\circ}\)

Answer.

Case 6; one triangle; \(a\approx1.61\text{,}\) \(A\approx16.1^{\circ}\text{,}\) \(B\approx43.9^{\circ}\)

20.

\(b=2\text{,}\) \(c=7\text{,}\) \(B=30^{\circ}\)

Answer.

Case 1; no triangles

21.

\(a=8\text{,}\) \(c=6\text{,}\) \(A=45^{\circ}\)

Answer.

Case 4; one triangle; \(b\approx11.02\text{,}\) \(B\approx103.0^{\circ}\text{,}\) \(C\approx32.0^{\circ}\)

22.

\(a=13\text{,}\) \(b=12\text{;}\) \(B=46^{\circ}\)

Answer.

Case 3; two triangles; \(c_1\approx16.55\text{,}\) \(A_1\approx51.2^{\circ}\text{,}\) \(C_1\approx82.8^{\circ}\text{;}\) \(c_2\approx1.51\text{,}\) \(A_2\approx128.8^{\circ}\text{,}\) \(C_2\approx5.2^{\circ}\)

23.

\(a=\sqrt{3}\text{,}\) \(c=2\text{,}\) \(A=60^{\circ}\)

Answer.

Case 2; one right triangle; \(b=1\text{,}\) \(B=30^{\circ}\text{,}\) \(C=90^{\circ}\)

24.

\(a=7\text{,}\) \(c=5\text{,}\) \(C=30^{\circ}\)

Answer.

Case 3; two triangles; \(b_1\approx9.63\text{,}\) \(A_1\approx44.4^{\circ}\text{,}\) \(B_1\approx105.6^{\circ}\text{;}\) \(b_2\approx2.49\text{,}\) \(A_2\approx135.6^{\circ}\text{,}\) \(B_2\approx14.4^{\circ}\)

25.

\(b=14\text{,}\) \(c=8\text{,}\) \(C=115^{\circ}\)

Answer.

Case 5; no triangles

26.

\(a=4\text{,}\) \(b=8\text{,}\) \(A=30^{\circ}\)

Answer.

Case 2; one right triangle; \(c\approx6.93\text{,}\) \(B=90^{\circ}\text{,}\) \(C=60^{\circ}\)

27.

\(b=12\text{,}\) \(c=4\text{,}\) \(C=55^{\circ}\)

Answer.

Case 1; no triangles

28.

\(a=7\text{,}\) \(b=10\text{,}\) \(B=108^{\circ}\)

Answer.

Case 6; one triangle; \(c\approx5.30\text{,}\) \(A\approx41.7^{\circ}\text{,}\) \(C\approx30.3^{\circ}\)

29.

\(a=4\text{,}\) \(c=5\text{,}\) \(A=100^{\circ}\)

Answer.

Case 5; no triangles

30.

\(b=7\text{,}\) \(c=4\text{,}\) \(B=35^{\circ}\)

Answer.

Case 4, one triangle; \(a\approx9.89\text{,}\) \(A\approx125.9^{\circ}\text{,}\) \(C\approx19.1^{\circ}\)

Exercise Group.

Find the area of each triangle. Round your answer to the nearest tenth.

31.

\(a = 8\text{,}\) \(b = 12\text{,}\) \(C = 60^{\circ}\text{.}\)

Answer.

\(41.6\)

32.

\(b = 10\text{,}\) \(c = 15\text{,}\) \(A = 45^{\circ}\text{.}\)

Answer.

\(53.0\)

33.

\(c = 6\text{,}\) \(a = 9\text{,}\) \(B = 30^{\circ}\text{.}\)

Answer.

\(13.5\)

34.

\(a = 5\text{,}\) \(c = 8\text{,}\) \(B = 45^{\circ}\text{.}\)

Answer.

\(14.1\)

35.

\(b = 7\text{,}\) \(c = 10\text{,}\) \(A = 60^{\circ}\text{.}\)

Answer.

\(30.3\)

36.

\(a = 12\text{,}\) \(b = 15\text{,}\) \(C = 75^{\circ}\text{.}\)

Answer.

\(86.9\)

37.

\(b = 6\text{,}\) \(c = 8\text{,}\) \(A = 30^{\circ}\text{.}\)

Answer.

\(12.0\)

38.

\(a = 9\text{,}\) \(c = 12\text{,}\) \(B = 60^{\circ}\text{.}\)

Answer.

\(46.8\)

39.

A waʻa is sailing in the direction Hikina (east) at 5 knots and spots an island in the house Lā Koʻolau (\(11.25^{\circ}\) above due east). Six hours later (30 nm) the navigator measures the island in the house Manu Hoʻolua (\(45^{\circ}\) above due west).

(a)

How far is the waʻa from the island at the time of its first measurement, rounded to the nearest nautical mile?

Answer.

26 nm

(b)

How far is the waʻa from the island at the time of its second measurement, rounded to the nearest nautical mile?

Answer.

7 nm

(c)

How close did the waʻa get to the island, rounded to the nearest nautical mile?

Hint.

Measure the altitude.

Answer.

5 nm

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