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# 5.5: General Solutions of Triangles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use the Pythagorean Theorem, trigonometry functions, the Law of Sines, and the Law of Cosines to solve various triangles.
• Understand when it is appropriate to use each method.
• Apply the methods above in real-world and applied problems.

In the previous sections we have discussed a number of methods for finding a missing side or angle in a triangle. Previously, we only knew how to do this in right triangles, but now we know how to find missing sides and angles in oblique triangles as well. By combining all of the methods we’ve learned up until this point, it is possible for us to find all missing sides and angles in any triangle we are given.

## Summary of Triangle Techniques

Below is a chart summarizing the triangle techniques that we have learned up to this point. This chart describes the type of triangle (either right or oblique), the given information, the appropriate technique to use, and what we can find using each technique.

Type of Triangle: Given Information: Technique: What we can find:
Right Two sides Pythagorean Theorem Third side
Right One angle and one side Trigonometric ratios Either of the other two sides
Right Two sides Trigonometric ratios Either of the other two angles
Oblique 2 angles and a non-included side (AAS) Law of Sines The other non-included side
Oblique 2 angles and the included side (ASA) Law of Sines Either of the non-included sides
Oblique 2 sides and the angle opposite one of those sides (SSA) – Ambiguous case Law of Sines The angle opposite the other side (can yield no, one, or two solutions)
Oblique 2 sides and the included angle (SAS) Law of Cosines The third side
Oblique 3 sides Law of Cosines Any of the three
angles

## Using the Law of Cosines

It is possible for us to completely solve a triangle using the Law of Cosines. In order to do this, we will need to apply the Law of Cosines multiple times to find all of the sides and/or angles we are missing.

Example 1: In triangle $ABC, a = 12, b = 13, c = 8$. Solve the triangle.

Solution: Since we are given all three sides in the triangle, we can use the Law of Cosines. Before we can solve the triangle, it is important to know what information we are missing. In this case, we do not know any of the angles, so we are solving for $\angle{A}, \angle{B}$, and $\angle{C}$. We will begin by finding $\angle{A}$.

$12^2 & = 8^2 + 13^2 - 2(8)(13) \cos A \\144 & = 233 - 208 \cos A \\- 89 & = - 208 \cos A \\0.4278846154 & = \cos A \\64.7 & \approx \angle{A}$

Now, we will find $\angle{B}$ by using the Law of Cosines. Keep in mind that you can now also use the Law of Sines to find $\angle{B}$. Use whatever method you feel more comfortable with.

$13^2 & = 8^2 + 12^2 - 2(8)(12) \cos B \\ 169 & = 208 - 192 \cos B \\ -39 & = -192 \cos B \\ 0.2031 & = \cos B \\ 78.3^\circ & \approx \angle{B}$

We can now quickly find $\angle{C}$ by using the Triangle Sum Theorem, $180^\circ - 64.7^\circ - 78.3^\circ = 37^\circ$

Example 2: In triangle $DEF, d = 43, e = 37$, and $\angle{F} = 124^\circ$. Solve the triangle.

Solution: In this triangle, we have the SAS case because we know two sides and the included angle. This means that we can use the Law of Cosines to solve the triangle. In order to solve this triangle, we need to find side $f,\angle{D}$, and $\angle{E}$. First, we will need to find side $f$ using the Law of Cosines.

$f^2 & = 43^2 + 37^2 - 2(43)(37) \cos 124 \\ f^2 & = 4997.351819 \\f& \approx 70.7$

Now that we know $f$, we know all three sides of the triangle. This means that we can use the Law of Cosines to find either angle $D$ or angle $E$. We will find angle $D$ first.

$43^2 & = 70.7^2 + 37^2 - 2(70.7)(37) \cos D \\1849 & = 6367.49 - 5231.8 \cos D \\-4518.49 & = -5231.8 \cos D \\0.863658779 & = \cos D \\ 30.3^\circ & \approx \angle{D}$

To find angle $E$, we need only to use the Triangle Sum Theorem, $\angle{E} = 180 - (124 + 30.3) = 25.7^\circ$.

Example 3: A control tower is receiving signals from two microchips implanted in wild tigers. Microchip 1 is 135 miles from the control tower and microchip 2 is 182 miles from the control tower. If the control tower forms a $119^\circ$ angle with both microchips, how far apart are the two tigers?

Solution: To find the distance between the two tigers, we need to find the distance between the two microchips. We will call this distance $x$. Since we know two sides and the included angle, we can use the Law of Cosines to find $x$.

$x^2 & = 135^2 + 182^2 - 2(135)(182) \cos 119 \\x^2 & = 75172.54474 \\x & = 274.2\ miles$

The two tigers are 274.2 miles apart.

## Using the Law of Sines

It is also possible for us to completely solve a triangle using the Law of Sines if we begin with the ASA case, the AAS case, or the SSA case. We must remember that when given the SSA case, it is possible that we may encounter the Ambiguous Case.

Example 4: In triangle $ABC, A = 43^\circ, B = 82^\circ$, and $c = 10.3$. Solve the triangle.

Solution: This is an example of the ASA case, which means that we can use the Law of Sines to solve the triangle. In order to use the Law of Sines, we must first know angle $C$, which we can find using the Triangle Sum Theorem, $\angle{C} = 180^\circ - (43^\circ + 82^\circ) = 55^\circ$.

Now that we know $\angle{C}$, we can use the Law of Sines to find either side $a$ or side $b$.

$\frac{\sin 55}{10.3} & = \frac{\sin 43}{a} && \frac{\sin 55}{10.3} = \frac{\sin 82}{b} \\a & = \frac{10.3 \sin 43}{\sin 55} && \qquad \ b = \frac{10.3 \sin 82}{\sin 55} \\a & = 8.6 && \qquad \ b = 12.5$

Example 5: A cruise ship is based at Island 1, but makes trips to Island 2 and Island 3 during the day. Island 3 lies directly east of Island 1. If the distance from Island 1 to Island 2 is 28.3 miles, from Island 2 to 3 is 52.4 miles, and Island 3 to 1 is 59.8 miles, what heading (angle) must the captain:

a. Leave Island 1

b. Leave Island 2

c. Leave Island 3

Solution: In order to find all three angles in the triangle, we must use the Law of Cosines because we are dealing with the SSS case. Once we find one angle using the Law of Cosines, we can use the Law of Sines to find a second angle. Then, we can use the Triangle Sum Theorem to find the third angle.

We will begin by finding $\angle{B}$ because it is the largest angle.

$59.8^2 & = 52.4^2 + 28.3^2 -2(52.4)(28.3) \cos B \\3576.04 & = 3546.65 - 2965.84 \cos B \\ 29.39 & = -2965.84 \cos B \\-0.0099095029 & = \cos B \\B & = 90.6^\circ$

Now that we know $\angle{B}$, we can find either $\angle{A}$ or $\angle{C}$. We will find $\angle{C}$ first since it is the second largest angle.

$\frac{\sin 90.6}{59.8} & = \frac{\sin C}{52.4} \\\frac{52.4 \sin 90.6}{59.8} & = \sin C \\0.876203135 & = \sin C \\\angle{C} & = 61.2^\circ$

Now that we know $\angle{B}$ and $\angle{C}$, we can use the Triangle Sum Theorem to find $\angle{A}= 180^\circ - (61.2^\circ + 90.6^\circ) = 28.2^\circ$.

Now, we must convert our angles into headings.

When going from Island 1 to Island 2, $61.2^\circ$ would be a heading of $N 28.8^\circ E$. Also, since $\angle{A}=28.2^\circ$, if we were to travel from Island 3 to Island 2, our heading would be $W 28.2^\circ N$. This means that when going from Island 2 to Island 3, the heading would be in the exact opposite direction, or $E 28.2^\circ S$. When going from Island 3 back to Island 1, since we know that Island 3 is directly east of Island 1, the captain must now sail in the direction opposite of east, or directly west, which is $N 90^\circ W$.

## Points to Consider

• Is there ever a situation where you would need to use the Law of Sines before using the Law of Cosines?
• In what situation might you consider using the Law of Cosines instead of Law of Sines if both were applicable?
• Why do we only have to use the Law of Cosines one time before we can switch to using the Law of Sines?

## Review Questions

1. Using the information provided, decide which case you are given (SSS, SAS, AAS, ASA, or SSA), and whether you would use the Law of Sines or the Law of Cosines to find the requested side or angle. Make an approximate drawing of each triangle and label the given information. Also, state how many solutions (if any) each triangle would have. If a triangle has no solution or two solutions, explain why.
Given Drawing Case Law Number of Solutions
a. $A = 69^\circ, B = 12^\circ, a = 22.3$, find $b$.
b. $a = 1.4, b = 2.3, C = 58^\circ$, find $c$.
c. $a = 3.3, b = 6.1, c = 4.8$, find $A$.
d. $a = 15, b = 25, A = 58^\circ$, find $B$.
e. $a = 45, b = 60, A = 47^\circ$, find $B$.
1. Using the information in the chart above, solve for the requested side or angle.
2. Using the information in the chart in question 1 and your answers from question 2, determine what information you are still missing from each triangle. Then, solve for each piece, solving each triangle.
3. The side of a rhombus is 12 cm and the longer diagonal is 21.5cm. Find the area of the rhombus and the measures of the angles in the rhombus.
4. Find the area of the pentagon below.
5. In the figure drawn below, angle $T$ is $56.8^\circ$ and $RT = 38$. Using the figure below, find the length of the altitude drawn to the longest side, the area of the two triangles formed by this altitude, $RI$ and angle $I$.
6. Refer back to Example 5, the island hopping problem. Suppose the cruise ship were based on Island 3 and traveled to Island 2 and then to Island 1, before returning to Island 3.
1. What would be its heading when going from Island 3 to Island 2?
2. What would be its heading when going from Island 2 to Island 1?
3. What would be its heading when going from Island 1 back to Island 3?
7. A golfer is standing on the tee of a golf hole that has a $115^\circ$ bend to the left. The distance from the tee to the bend is 218 yards. The distance from the bend to the green is 187 yards.
1. How far would the golfer need to hit the ball if he wanted to make it to the green in one shot?
2. At what angle would he need to hit the ball?
8. A golfer is standing on the tee, which is 320 yards from the cup on the green. After he hits his first shot, which is sliced to the right, his ball forms a $162.2^\circ$ angle with the tee and the cup, and the cup forms a $14.2^\circ$ angle with his ball and the tee.
1. What is the degree of his slice?
2. How far was his first shot?
3. How far away from the cup is he?

1. AAS, Law of Sines, one solution
2. SAS, Law of Cosines, one solution
3. SSS, Law of Cosines, one solution
4. SSA, Law of Sines, no solution $(15 < 25 \sin 58^\circ)$
5. SSA, Law of Sines, two solutions $(45 > 60 \sin 47^\circ)$
1. $\frac{\sin 69^\circ}{22.3} = \frac{\sin 12^\circ}{b}, b = 4.97$
2. $c^2 = 1.4^2 + 2.3^2 - 2(1.4)(2.3)\cos 58^\circ, c = 2.0$
3. $3.3^2 = 6.1^2 + 4.8^2 -2(6.1)(4.8) \cos A, A = 32.6^\circ$
4. No solution
5. $\frac{\sin 47^\circ}{45} = \frac{\sin B}{60}, B = 77.2^\circ$ or $180^\circ - 77.2^\circ = 102.8^\circ$
1. (a) need angle $C$ and side $c$. $C & = 180^\circ - 69^\circ - 12^\circ = 99^\circ \\ \frac{\sin 99^\circ}{c} & = \frac{\sin 69^\circ}{22.3}, c = 23.6$ (b) need angle $A$ and angle $B$. $\frac{\sin 58^\circ}{2} = \frac{\sin A}{1.4}, A & = 36.4^\circ \\ 180^\circ - 36.4^\circ - 58^\circ, B & = 85.6^\circ$ (c) need angle $B$ and angle $C$. $\frac{\sin32.6^\circ}{3.3} = \frac{\sin B}{6.1}, B & = 84.8^\circ \\180^\circ - 32.6^\circ - 84.8^\circ, C & = 62.6^\circ$ (d) No solution (e) Both cases need angle $C$ and side $c$. $\text{Case}\ 1: C & = 180^\circ - 47^\circ - 77.2^\circ = 55.8^\circ, \frac{\sin 55.8^\circ}{c} = \frac{\sin 47^\circ}{45}, c = 50.9 \\\text{Case}\ 2: C & = 180^\circ - 47^\circ - 102.8^\circ = 30.2^\circ,\frac{\sin 30.2^\circ}{c} = \frac{\sin 47^\circ}{45}, c = 30.95$
2. To find the area of the rhombus, use the formula $K = \frac{1}{2} \ bc \sin A$ and then multiply that by 2. We first need to find one of the angles that are opposite the given diagonal (they are both the same measurement). We will call it angle $A$. $21.5^2 = 12^2 + 12^2 - 2(12)(12) \sin A, A = 127.2^\circ$, which means the other two angles are both $52.8^\circ (360^\circ - 127.2^\circ - 127.2^\circ$ and then divide by 2). $K = 2 \left (\frac{1}{2}(12)(12) \sin 127.2^\circ \right ) = 114.7$
3. Divide the pentagon into three triangles, drawing segments from $\angle{2}$ to $\angle{5}$, called $x$ below, and $\angle{2}$ to $\angle{4}$, called $y$ below. With these three triangles, only the middle triangle needs us to find two sides and the angle between them (called $\angle{Z}$ below) to use $K = \frac{1}{2} \ bc \sin A$ (the outer two triangles already have two sides and an angle that fit this criteria). $x^2 & = 192^2 + 190.5^2 - 2(192)(190.5)\cos 81^\circ \rightarrow x = 248.4 \\y^2 & = 146^2 + 173.8^2 - 2(146)(173.8)\cos 73^\circ \rightarrow x = 191.5 \\118^2 & = 248.4^2 + 191.5^2 - 2(248.4)(191.5)\cos Z \rightarrow \angle{Z} = 27.4^\circ$ Areas: $K = \frac{1}{2}(190.5)(192)\sin 81^\circ = 18062.8$ $k & = \frac{1}{2}(248.4)(191.5)\sin 27.4^\circ = 10945.5 \\K & = \frac{1}{2}(173.8)(146)\sin 73^\circ = 12133.0 && \text{Total Area}:\ 41141.3$
4. altitude, $x$: $\sin 56.8^\circ = \frac{x}{38} \rightarrow x = 31.8$ $GT & = \sqrt{38^2 - 31.8^2} = 20.8 && GI = 88 - 20.8 = 67.2\\A_{small} & = \frac{1}{2} (20.8)(31.8) = 330.8 && A_{big} = \frac{1}{2}(67.2)(31.8) = 1068.5 \\RI & = \sqrt{67.2^2 + 31.8^2} = 74.3 && \angle{I} \rightarrow \sin I = \frac{31.8}{74.3} \rightarrow 25.3^\circ$
5. The headings would be as follows:
1. $W 28.2^\circ N$
2. $S 28.8^\circ W$
3. $N 90^\circ E$
1. $a^2 = 187^2 + 218^2 - 2(187)(218)\cos 115^\circ$, he would need to hit the ball 342.0 yards.
2. $\frac{\sin 115^\circ}{342} = \frac{\sin B}{187}$, he would have to hit the ball at a $29.7^\circ$ angle.
1. $180^\circ - 14.2^\circ - 162.2^\circ = 3.6^\circ$
2. $\frac{\sin 14.2^\circ}{b} = \frac{\sin 162.2^\circ}{320}, 256.8\ yards$
3. $\frac{\sin 162.2^\circ}{320} = \frac{\sin 3.6^\circ}{c}, 65.7\ yards$

Feb 23, 2012

Aug 06, 2015