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8.2 Non-right triangles: law of cosines  (Page 3/8)

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Using the law of cosines to solve a communication problem

On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.

For simplicity, we start by drawing a diagram similar to [link] and labeling our given information.

A triangle formed between the two cell phone towers located on am east to west highway and the cellphone between and north of them. The side between the two towers is 6000 feet, the side between the left tower and the phone is 5050 feet, and the side between the right tower and the phone is 2420 feet. The angle between the 5050 and 6000 feet sides is labeled theta.

Using the Law of Cosines, we can solve for the angle   θ .   Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let   a = 2420 , b = 5050 ,   and   c = 6000.   Thus,   θ   corresponds to the opposite side   a = 2420.  

                                                 a 2 = b 2 + c 2 − 2 b c cos   θ                                        ( 2420 ) 2 = ( 5050 ) 2 + ( 6000 ) 2 − 2 ( 5050 ) ( 6000 ) cos   θ             ( 2420 ) 2 − ( 5050 ) 2 − ( 6000 ) 2 = − 2 ( 5050 ) ( 6000 ) cos   θ     ( 2420 ) 2 − ( 5050 ) 2 − ( 6000 ) 2 − 2 ( 5050 ) ( 6000 ) = cos   θ                                            cos   θ ≈ 0.9183                                                  θ ≈ cos − 1 ( 0.9183 )                                                  θ ≈ 23.3°

To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in [link] . This forms two right triangles, although we only need the right triangle that includes the first tower for this problem.

The triangle between the phone, the left tower, and a point between the phone and the highway between the towers. The side between the phone and the highway is perpendicular to the highway and is y feet. The highway side is x feet. The angle at the tower, previously labeled theta, is 23.3 degrees.

Using the angle   θ = 23.3°   and the basic trigonometric identities, we can find the solutions. Thus

              cos ( 23.3° ) = x 5050                     x = 5050 cos ( 23.3° )                     x ≈ 4638.15   feet      sin ( 23.3° ) = y 5050                     y = 5050 sin ( 23.3° )                     y ≈ 1997.5   feet

The cell phone is approximately 4638 feet east and 1998 feet north of the first tower, and 1998 feet from the highway.

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Calculating distance traveled using a sas triangle

Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in [link] .

A triangle whose vertices are the boat, the port, and the turning point of the boat. The side between the port and the turning point is 10 mi, and the side between the turning point and the boat is 8 miles. The side between the port and the turning point is extended in a straight dotted line. The angle between the dotted line and the 8 mile side is 20 degrees.

The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, 180° − 20° = 160° .   With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the boat to the port.

  x 2 = 8 2 + 10 2 − 2 ( 8 ) ( 10 ) cos ( 160° )   x 2 = 314.35         x = 314.35         x ≈ 17.7   miles

The boat is about 17.7 miles from port.

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Using heron’s formula to find the area of a triangle

We already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.

Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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