# Chapter 5: Triangles and Vectors

**At Grade**Created by: CK-12

- 5.1.
## The Law of Cosines

- 5.2.
## Area of a Triangle

- 5.3.
## The Law of Sines

- 5.4.
## The Ambiguous Case

- 5.5.
## General Solutions of Triangles

- 5.6.
## Vectors

- 5.7.
## Component Vectors

### 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 \begin{align*}\frac{1}{2} \ bc \sin A\end{align*}

## 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. \begin{align*}K = \sqrt{s(s - a)(s - b)(s - c)}\end{align*}
K=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√ where \begin{align*}s =\frac{1}{2}(a+b+c)\end{align*}s=12(a+b+c) 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. \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*}
a2=b2+c2−2bccosA or b2=a2+c2−2accosB or c2=a2+b2−2abcosC

- Law of Sines
- A statement about the relationship between the sides and the angles in any triangle. \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*}
sinAa=sinBb=sinCc or asinA=bsinB=csinC

- 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

- Use the Law of Cosines to determine whether or not the following triangle is drawn accurately. If not, determine how much \begin{align*}\angle{B}\end{align*}
∠B is off by. - 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 \begin{align*}BC = 51.4\ feet, BD = 32.6\ feet, AD = 37.3\ feet\end{align*}
BC=51.4 feet,BD=32.6 feet,AD=37.3 feet and \begin{align*}\angle{DBC} = 27^\circ\end{align*}∠DBC=27∘ , verify that the length of \begin{align*}AB\end{align*}AB would be 34.3 feet. If not, figure out the correct measure. - 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?
- 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 \begin{align*}E\end{align*}
E . 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 \begin{align*}A\end{align*}A and to a point on the lower arm 4 cm from the mechanical elbow at point \begin{align*}B\end{align*}B . When fluid is pumped out of the cylinder the distance \begin{align*}AB\end{align*}AB is shortened. The forearm goes up raising the hand at point \begin{align*}H\end{align*}H . Some fluid is pumped out of the cylinder to make the distance \begin{align*}AB\end{align*}AB 5 cm shorter. What is the new angle of the arm, \begin{align*}\angle{AEH}\end{align*}∠AEH ? - For each figure below, use the Law of Sines, the Law of Cosines, or the the trig functions to solve for \begin{align*}x\end{align*}
x . - 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.
- If the first angle formed is \begin{align*}29^\circ\end{align*}
29∘ and the second angle formed is \begin{align*}3^\circ\end{align*}3∘ , find the distance across the canyon. - The surveyor spots another boulder while he is at his second spot, and finds that it forms a \begin{align*}37^\circ\end{align*}
37∘ angle with his line of sight. He then drives 15 km further and finds that the boulder forms a \begin{align*}25^\circ\end{align*}25∘ angle with the line of sight. What is the distance between the two boulders?

- If the first angle formed is \begin{align*}29^\circ\end{align*}
- 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 \begin{align*}72.8^\circ\end{align*}
72.8∘ 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 \begin{align*}12^\circ\end{align*}12∘ 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? - Find the magnitude and direction of each resultant vector (addition). \begin{align*}\vec{m}\end{align*}
m⃗ and \begin{align*}\vec{n}\end{align*}n⃗ are perpendicular.- \begin{align*}|\vec{m}| = 48.3,|\vec{n}| = 47.6\end{align*}
|m⃗|=48.3,|n⃗|=47.6 - \begin{align*}|\vec{m}| = 18.6,|\vec{n}| = 17.5\end{align*}
|m⃗|=18.6,|n⃗|=17.5

- \begin{align*}|\vec{m}| = 48.3,|\vec{n}| = 47.6\end{align*}
- Given the initial and terminal coordinates of \begin{align*}\vec{a}\end{align*}
a⃗ , find the magnitude and direction.- initial (-4, 19) terminal (12, 1)
- initial (11, -21) terminal (21, -11)

- A golfer tees off at the \begin{align*}16^{th}\end{align*}
16th hole. The pin is 425 yards from tee-off and his ball is \begin{align*}16^\circ\end{align*}16∘ off of the straight line to the hole. If his ball is 137 yards from the hole, how far did he hit the ball? - 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 \begin{align*}36^\circ\end{align*}
36∘ 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? - During a baseball game, a ball is hit into right field. The angle from the ball to home to \begin{align*}2^{nd}\end{align*}
2nd base is \begin{align*}18^\circ\end{align*}18∘ . The angle from the ball to \begin{align*}2^{nd}\end{align*}2nd to home is \begin{align*}127^\circ\end{align*}127∘ . The distance from home to \begin{align*}2^{nd}\end{align*}2nd base is 127.3 ft. How far was the ball hit? How far is second base from the ball? - 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?
- 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 \begin{align*}5.2 \times 10^{13}\end{align*}
5.2×1013 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?

## Texas Instruments Resources

*In the CK-12 Texas Instruments Trigonometry FlexBook® resource, 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.*

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