# 4.2 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.

Using the Law of Cosines, we can solve for the angle $\text{\hspace{0.17em}}\theta .\text{\hspace{0.17em}}$ Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let $\text{\hspace{0.17em}}a=2420,b=5050,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}c=6000.\text{\hspace{0.17em}}$ Thus, $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ corresponds to the opposite side $\text{\hspace{0.17em}}a=2420.\text{\hspace{0.17em}}$

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.

Using the angle $\text{\hspace{0.17em}}\theta =23.3°\text{\hspace{0.17em}}$ and the basic trigonometric identities, we can find the solutions. Thus

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

## 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] .

The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, $180°-20°=160°.\text{\hspace{0.17em}}$ 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.

$\begin{array}{l}\text{\hspace{0.17em}}{x}^{2}={8}^{2}+{10}^{2}-2\left(8\right)\left(10\right)\mathrm{cos}\left(160°\right)\hfill \\ \text{\hspace{0.17em}}{x}^{2}=314.35\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\sqrt{314.35}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\approx 17.7\text{\hspace{0.17em}}\text{miles}\hfill \end{array}$

The boat is about 17.7 miles from port.

## 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.

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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Commplementary angles
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Sherica
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Sherica
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Tamia
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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J, combine like terms 7x-4y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Cied
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Porter
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Stotaw
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Azam
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Azam
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Prasenjit
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Damian
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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