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# Chapter 5: Triangles and Vectors

Difficulty Level: At Grade Created by: CK-12

## Chapter Summary

This chapter has taught us how to solve any kind of triangle, using the Law of Cosines, Law of Sines and vectors. We also discovered two additional formulas for finding the area, Heron’s Formula and 12 bcsinA\begin{align*}\frac{1}{2} \ bc \sin A\end{align*}. Then, the ambiguous case for the Law of Sines was introduced. This is when you are given two sides and the non-included angle and have to solve for another angle in the triangle. There can be no, one or two solutions and you need to compare the two given sides to determine which option it is. Finally, vectors were discussed. We learned how to add, subtract and multiply them by a scalar. Vectors are very useful for representing speed, wind velocity and force.

## Vocabulary

Ambiguous case
A situation that occurs when applying the Law of Sines in an oblique triangle when two sides and a non-included angle are known. The ambiguous case can yield no solution, one solution, or two solutions.
component vectors
Two or more vectors whose vector sum, the resultant, is the given vector. Components can be on axes or more generally in space.
directed line segment
A line segment having both magnitude and direction, often used to represent a vector.
displacement
When an object moves a certain distance in a certain direction.
equal vectors
Vectors with the same magnitude and direction.
force
When an object is pushed or pulled in a certain direction.
Heron’s Formula
A formula used to calculate the area of a triangle when all three sides are known. K=s(sa)(sb)(sc)\begin{align*}K = \sqrt{s(s - a)(s - b)(s - c)}\end{align*} where s=12(a+b+c)\begin{align*}s =\frac{1}{2}(a+b+c)\end{align*} or half of the perimeter of the triangle.
included angle
The angle in between two known sides of a triangle.
included side
The side in between two known sides of a triangle.
initial point
The starting point of a vector
Law of Cosines
A general statement relating the lengths of the sides of a general triangle to the cosine of one of its angles. a2=b2+c22bccosA or b2=a2+c22accosB or c2=a2+b22abcosC\begin{align*}a^2 = b^2 + c^2 - 2bc \cos A \ \text{or}\ b^2 = a^2 + c^2 - 2ac \cos B \ \text{or}\ c^2 = a^2 + b^2 - 2ab \cos C\end{align*}
Law of Sines
A statement about the relationship between the sides and the angles in any triangle. sinAa=sinBb=sinCc or asinA=bsinB=csinC\begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\ \text{or}\ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\end{align*}
magnitude
Length of a vector.
negative vector
A vector with the same magnitude as the original vector but with the opposite direction.
non-included angle
An angle that is not in between two known sides of a triangle.
non-included side
A side that is not in between two known sides of a triangle.
oblique triangle
A non-right triangle.
resultant
The sum of two or more vectors
scalar
A real number by which a vector can be multiplied. The magnitude of a vector is always a scalar.
standard position
A vector with its initial point at the origin of a coordinate plane.
terminal point
The ending point of a vector.
unit vector
A vector that has a magnitude of one unit. These generally point on coordinate axes.
vector
Any quantity having magnitude and direction, often represented by an arrow.
velocity
When an object travels at a certain speed in a certain direction.

## Review Questions

1. Use the Law of Cosines to determine whether or not the following triangle is drawn accurately. If not, determine how much B\begin{align*}\angle{B}\end{align*} is off by.
2. An artist is making a large sculpture for the lobby of a new building. She has drawn out what she wants the sculpture to look like at the left. If she wants BC=51.4 feet,BD=32.6 feet,AD=37.3 feet\begin{align*}BC = 51.4\ feet, BD = 32.6\ feet, AD = 37.3\ feet\end{align*} and DBC=27\begin{align*}\angle{DBC} = 27^\circ\end{align*}, verify that the length of AB\begin{align*}AB\end{align*} would be 34.3 feet. If not, figure out the correct measure.
3. A family’s farm plot is a trapezoid with dimensions: the longer leg is 3,000 ft and the shortest side is 2,100 ft. The other leg is 2,400 ft. The shorter diagonal is 2,200 ft. What is the area of the land in square feet?
4. A biomechanics class is designing a functioning artificial arm for an adult. They are using a hydraulic cylinder (fluid filled) to be the bicep’s muscle. The elbow is at point E\begin{align*}E\end{align*}. The forearm dimension EH is 24 cm. The upper arm dimension EA is 21 cm. The cylinder attaches from the top of the upper arm at point A\begin{align*}A\end{align*} and to a point on the lower arm 4 cm from the mechanical elbow at point B\begin{align*}B\end{align*}. When fluid is pumped out of the cylinder the distance AB\begin{align*}AB\end{align*} is shortened. The forearm goes up raising the hand at point H\begin{align*}H\end{align*}. Some fluid is pumped out of the cylinder to make the distance AB\begin{align*}AB\end{align*} 5 cm shorter. What is the new angle of the arm, AEH\begin{align*}\angle{AEH}\end{align*}?
5. For each figure below, use the Law of Sines, the Law of Cosines, or the the trig functions to solve for x\begin{align*}x\end{align*}.
6. A surveyor has the job of determining the distance across the Palo Duro Canyon in Amarillo, Texas, the second largest canyon in the United States. Standing on one side of the canyon, he measures the angle formed by the edge of the canyon and the line of sight to a large boulder on the other side of the canyon. He then drives 12 km and measures the angle formed by the edge of the canyon and the new line of sight to the boulder.
1. If the first angle formed is 29\begin{align*}29^\circ\end{align*} and the second angle formed is 3\begin{align*}3^\circ\end{align*}, find the distance across the canyon.
2. The surveyor spots another boulder while he is at his second spot, and finds that it forms a 37\begin{align*}37^\circ\end{align*} angle with his line of sight. He then drives 15 km further and finds that the boulder forms a 25\begin{align*}25^\circ\end{align*} angle with the line of sight. What is the distance between the two boulders?
7. Two cell phone companies have towers along Route 47. Company 1’s tower is 38 miles from one point on Route 47 and 47 miles from another point. This tower’s signal forms a 72.8\begin{align*}72.8^\circ\end{align*} angle. Company 2’s tower is 58 miles from one point of Route 47 and 59 miles from another. Company 2’s signal forms a 12\begin{align*}12^\circ\end{align*} angle with the road at the point that is 58 miles from the tower. For how many miles would a person driving along Route 47 have service with Company A? Company B? If the signals start to overlap 32 miles from Tower 1 (along the same line as the 47 mile), how long does the two coverages overlap over the highway?
8. Find the magnitude and direction of each resultant vector (addition). m\begin{align*}\vec{m}\end{align*} and n\begin{align*}\vec{n}\end{align*} are perpendicular.
1. |m|=48.3,|n|=47.6\begin{align*}|\vec{m}| = 48.3,|\vec{n}| = 47.6\end{align*}
2. |m|=18.6,|n|=17.5\begin{align*}|\vec{m}| = 18.6,|\vec{n}| = 17.5\end{align*}
9. Given the initial and terminal coordinates of a\begin{align*}\vec{a}\end{align*}, find the magnitude and direction.
1. initial (-4, 19) terminal (12, 1)
2. initial (11, -21) terminal (21, -11)
10. A golfer tees off at the 16th\begin{align*}16^{th}\end{align*} hole. The pin is 425 yards from tee-off and his ball is 16\begin{align*}16^\circ\end{align*} off of the straight line to the hole. If his ball is 137 yards from the hole, how far did he hit the ball?
11. Street A runs north and south and intersects with Street B, which runs east and west. Street C intersects both A and B, and it intersects Street A at a 36\begin{align*}36^\circ\end{align*} angle. There are stoplights at each of these intersections. If the distance between the two stoplights on Street C is 0.5miles, what is the distance between the two stoplights on Street A?
12. During a baseball game, a ball is hit into right field. The angle from the ball to home to 2nd\begin{align*}2^{nd}\end{align*} base is 18\begin{align*}18^\circ\end{align*}. The angle from the ball to 2nd\begin{align*}2^{nd}\end{align*} to home is 127\begin{align*}127^\circ\end{align*}. The distance from home to 2nd\begin{align*}2^{nd}\end{align*} base is 127.3 ft. How far was the ball hit? How far is second base from the ball?
13. The military is testing out a new infrared sensor that can detect movement up to thirty miles away. Will the sensor be able to detect the second target? If not, how far out of the range of the sensor is Target 2?
14. An environmentalist is sampling the water in a local lake and finds a strain of bacteria that lives on the surface of the lake. In a one square foot area, he found 5.2×1013\begin{align*}5.2 \times 10^{13}\end{align*} bacteria. There are three docks in a certain section of the lake. If Dock 3 is 396 ft from Dock 1, how many bacteria are living on the surface of the water between the three docks?

1. Use Law of Cosines to solve for angle B\begin{align*}B\end{align*}. 17.62=152+20.922(15)(20.9)cosB,B=55.8\begin{align*}17.6^2 = 15^2 + 20.9^2 - 2 (15)(20.9)\cos B, B = 55.8^\circ\end{align*}, which means the picture was drawn accurately.
2. CD2C, CAB2=32.62+51.422(32.6)(51.4)cos27,CD=26.8=33.5=64.12+51.422(64.1)(51.4)cos33.5,AB=35.4
This means that the artist’s drawing is off by 1.1 foot.
3. The area of a trapezoid is A=12h(b1+b2)\begin{align*}A = \frac{1}{2}h(b_1 + b_2)\end{align*}. In the problem, we were given both bases, a side and a diagonal. So, we need to find the height. In order to do this, we first need to find the angle between 2200 and 2400(x)\begin{align*}2400 (x)\end{align*}, use that to find it’s complement, then we can take the cosine to find the height.
21002xcos35.9A=24002+220022(2400)(2200)cosx=54.19054.1=35.9=h2200h=1782.1=12(1782.1)(2400+3000)=4,811,670 sq. ft.
4. AB2=42+2122(4)(21)cos120,AB=23.3 minus 5\begin{align*}AB^2 = 4^2 + 21^2 - 2(4)(21)\cos 120, AB = 23.3\ minus\ 5\end{align*}, new AB\begin{align*}AB\end{align*} is 18.3 new E\begin{align*}\angle{E}\end{align*} is 18.32=42+2122(4)(21)cosE,E=42.6\begin{align*}18.3^2 = 4^2 + 21^2 - 2(4)(21)\cos E,\angle{E} = 42.6^\circ\end{align*}
1. x2=13.32+17.622(13.3)(17.6)cos132,x=28.3\begin{align*}x^2 = 13.3^2 + 17.6^2 - 2(13.3)(17.6)\cos 132, x = 28.3\end{align*}
2. cos52=x19.3,x=11.9\begin{align*}\cos 52 = \frac{x}{19.3}, x = 11.9\end{align*}
3. sin4218.6=sinx15.9,x=34.9\begin{align*}\frac{\sin 42}{18.6} = \frac{\sin x}{15.9}, x = 34.9^\circ\end{align*}
4. sin4692.7=sin12x,x=26.8\begin{align*}\frac{\sin 46}{92.7} = \frac{\sin 12}{x}, x = 26.8\end{align*}
1. We will call the distance across the canyon h\begin{align*}h\end{align*}. sin3212=sin87x,x=22.6.\begin{align*}\frac{\sin 32}{12} = \frac{\sin 87}{x}, x = 22.6.\end{align*} sin61=h22.6,19.8\begin{align*}\sin 61 = \frac{h}{22.6}, 19.8\end{align*} km.
2. We will call the distance between the two boulders y\begin{align*}y\end{align*}. y=a+b\begin{align*}y = a + b\end{align*}, where a\begin{align*}a\end{align*} is the distance from boulder 1 to the first stop (that the surveyor took) and b\begin{align*}b\end{align*} is the distance from that spot to boulder 2.

tan3=a19.8tan37=b19.8
So, a+b=19.8(tan3+tan37)=16.0\begin{align*}a + b = 19.8(\tan 3^\circ + \tan 37^\circ) = 16.0\end{align*} km.

5. Company A coverage: x2=382+4722(38)(47)cos72.8,x=51\begin{align*}x^2 = 38^2 + 47^2 - 2(38)(47)\cos 72.8, x = 51\end{align*} Company B coverage: sinx58=sin1259,x=11.8sin1259=sin156.2b,b=114.52\begin{align*}\frac{\sin x}{58} = \frac{\sin 12}{59}, x = 11.8^\circ \rightarrow \frac{\sin 12}{59} = \frac{\sin 156.2}{b}, b = 114.52 \end{align*} Overlap: sina38=sin72.851,a=45.4sin1215=sin122.6o,o=60.8\begin{align*}\frac{\sin a}{38} = \frac{\sin 72.8}{51}, a = 45.4^\circ \rightarrow \frac{\sin 12}{15} = \frac{\sin 122.6}{o}, o = 60.8\end{align*}
1. magnitude =48.32+47.62=67.8\begin{align*}=\sqrt{48.3^2 + 47.6^2} = 67.8\end{align*}, direction =tanθ=47.648.344.6\begin{align*}= \tan \theta = \frac{47.6}{48.3} \rightarrow 44.6^\circ\end{align*}
2. magnitude =18.62+17.52=25.5\begin{align*}=\sqrt{18.6^2 +17.5^2} = 25.5\end{align*}, direction =tanθ=47.648.343.3\begin{align*}=\tan \theta = \frac{47.6}{48.3} \rightarrow 43.3^\circ\end{align*}
1. magnitude =(191)2+(412)2=24.1\begin{align*}=\sqrt{(19-1)^2 + (-4-12)^2} = 24.1\end{align*}, direction =tanθ=181648.4\begin{align*}=\tan \theta = \frac{-18}{16} \rightarrow -48.4^\circ\end{align*}
2. magnitude =(21+11)2+(1121)2=14.1\begin{align*}=\sqrt{(-21+11)^2 +(11-21)^2} =14.1\end{align*}, direction =tanθ=101045\begin{align*}=\tan \theta = \frac{-10}{-10} \rightarrow 45^\circ\end{align*}
6. This is the SSA or ambiguous case. Because 137>425sin16\begin{align*}137 > 425 \sin 16^\circ\end{align*}, we will have two solutions or two different lengths that the golfer could have hit the ball. sin16137=sinx425,x=58.8\begin{align*}\frac{\sin 16}{137} = \frac{\sin x}{425}, x = 58.8^\circ\end{align*} or 121.2\begin{align*}121.2^\circ\end{align*} (this is the angle at the ball) Case 1: Use 58.8\begin{align*}58.8^\circ\end{align*}, the angle at the pin is 105.2\begin{align*}105.2^\circ\end{align*} sin105.2d=sin16137\begin{align*}\frac{\sin 105.2}{d} = \frac{\sin 16}{137}\end{align*}, distance is 479.6 yards. Case 2: Use 121.2\begin{align*}121.2^\circ\end{align*}, the angle at the pin is 42.8\begin{align*}42.8^\circ\end{align*} sin42.8d=sin16137\begin{align*}\frac{\sin 42.8}{d} = \frac{\sin 16}{137}\end{align*}, distance is 337.7 yards. It is safe to say that the golfer did not hit the ball 479.6 yards, considering that Tiger Woods longest recorded drive is 425 yards. So, we can logically rule out Case 1 and our answer is that the golfer’s drive was 337.7 yards.
7. cos36=x0.5,x=0.4 miles\begin{align*}\cos 36^\circ = \frac{x}{0.5}, x = 0.4\ miles\end{align*}
8. The third angle in the triangle is 35\begin{align*}35^\circ\end{align*}, from the Triangle Sum Theorem.
sin35127.3=sin127x,x=177.2
Therefore, the ball was hit 177.2 feet. The distance between second base and the ball can be calculated as follows:
sin35127.3=sin18x,x=68.6
9. Therefore, the distance from second base to the ball is 68.6 feet.
10. The distance between the tower and target 2 is: x2=372+1822(37)(18)cos67.2,34.3 miles\begin{align*}x^2 = 37^2 + 18^2 - 2(37)(18) \cos 67.2^\circ, 34.3\ miles\end{align*}. This means that the second target is out of range by 4.3 miles.
11. We need to find area, use Heron’s Formula. \begin{align*}s = \frac{1}{2}(587 + 247+396) = 615\end{align*}

\begin{align*}A = \sqrt{615(615-587)(615-247)(615-396)} = 37, 253.1\end{align*}, times \begin{align*}5.2 \times 10^{13} = 1.9 \times 10^{18}\end{align*} bacteria.

## Texas Instruments Resources

In the CK-12 Texas Instruments Trigonometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9703.

Sep 26, 2013