4 questions mostly finished and just need to be checked pls

[r75x]

Hello,

#1. An observer notes that the angle of elevation from poitn A to the top of a space shuttle is 27.2 degrees. From a point 17.5 meters further from the space shuttle, the angle of elevation is 23.9 degrees. Find the height of the space shuttle.

My work:

tan 23.9 = h/(x+17.5)
tan 27.2 = h/x
h = xtan27.2

h = xtan23.9 +17.5tan23.9
xtan27.2 = xtan23.9 + 17.5tan23.9
xtan27.2 - xtan23.9 = 17.5tan23.9
[x(tan27.2 - tan23.9)] = [17.5tan23.9]
[x(tan27.2 - tan23.9)]/[tan27.2 - tan 23.9] = [17.5tan23.9]/[tan27.2 - tan 23.9]
x = 109.55m

h = 109.55tan27.2 = 56.30m

The shuttle is 56.30m, correct?.

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#2. A submarine traveling 9mph is descending at an angle of depression of 5 degrees. How many minutes does it take the submarine to reach a depth of 80 feet?

My work:

distance = rate x time

80ft = [(9miles/1hour)(1hour/60min)(5280ft/1mile)]T
80ft = [792ft/min]T
T = 0.1010min

Do I do 0.1010tan(5*) now? I get 0.0088, but I don't think that makes sense. It'll take 0.0088 minutes for the submarine to reach a depth of 80 feet?

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#3. A car with a tire radius 15in is rotating at 450 revolutiosn per minute. Find the speed of the automobile to the nearest mile per hour.

My work:

450revolutions/minute = 900pi radians/minute

(15in)(1ft/12in)(1mi/5280ft) = 2.3674 x 10^-4 miles

(900pi radians/min)(60min/1hour) = 54000pi radians/hour

(2.3674 x 10^-4)(54,000pi) = 40.2mph

Is the car going 40.2mph?

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#4. Prove that cos(3x) + cos(x) = 4(cos^3)x - 2cos(x)

My work:

I'm not very good at proofs. I looked at this and I can't figure out which cosines to change around to make one side equal the other.

I know that I can use these identities to work with:

cosx = 1/secx
cos^2x = 1 - (sin^2)x

I can't really see anything working out though, probably because I'm still new at this...

Thank you for checking my work and helping me out!

 

[soroban]

Re: 4 questions mostly finished and just needs to be checked

Hello, r75x!

Your work for #1 is excellent and correct . . . beautiful work!

#2. A submarine traveling 9mph is descending at an angle of depression of 5 degrees.
How many minutes does it take the submarine to reach a depth of 80 feet?

First, we must find the distance the sub will travel . . . along the hypotenuse.

In the right triangle: \(\displaystyle \,\sin5^o\:=\:\frac{80}{d}\;\;\Rightarrow\;\;d\:=\:\frac{80}{\sin5^o} \:\approx\:917.9\) feet.

At 792 ft/min, it will take: \(\displaystyle \,\frac{917.9}{792} \:=\:1.16\) minutes.

#3. A car with a tire radius 15 inches is rotating at 450 rev/min.
Find the speed of the automobile to the nearest mile per hour.

I wouldn't bother with radians . . .
The diameter is 30 inches = 2.5 feet.
The circumference of the tire is: \(\displaystyle \,2.5\pi\) feet.
\(\displaystyle \;\;\)So: \(\displaystyle \,1 \text{ rev}\:=\:2.5\pi\) feet.

So we have: \(\displaystyle \,\frac{450\text{ rev}}{1\text{ min}}\,\times\,\frac{1\text{ min}}{60\text{ sec}} \,\times\,\frac{2.5\pi\text{ ft}}{1\text{ rev}}\,\times\,\frac{60\text{ mi/hr}}{88\text{ ft/sec}}\;\approx\;40\) miles per hour.

But there's absolutely nothing wrong with your approach . . . nice work!

#4. Prove that: \(\displaystyle \:\cos(3x)\,+\,\cos(x)\:=\:4\cdot\cos^3x\,-\,2\cdot\cos(x)\)

You'll need an Addition Formulas for this one:
\(\displaystyle \;\;\cos(A\,+\,B)\;=\;\cos(A)\cdot\cos(B)\,-\,\sin(A)\cdot\sin(B)\)

and some Double-angle Formulas:
\(\displaystyle \;\;\sin(2A)\;=\;2\cdot\sin(A)\cdot\cos(A)\)
\(\displaystyle \;\;\cos(2A)\;=\;2\cdot\cos^2(A)\,-\,1\)

and "invent" a formula for \(\displaystyle \cos(3x)\) . . .


We have: \(\displaystyle \,\cos(3x) \;= \;\cos(2x\,+\,x)\)

\(\displaystyle \;\;\;=\;\;\;\underbrace{\cos(2x)}\cdot\cos(x)\;\;-\;\;\underbrace{\sin(2x)}\cdot\sin(x)\)

\(\displaystyle \;\;\;=\;[\overbrace{2\cos^2(x)\,-\,1}]\cdot\cos(x) \,-\,[\overbrace{2\cdot\sin(x)\cdot\cos(x)}]\cdot\sin(x)\)

\(\displaystyle \;\;\;=\;2\cdot\cos^3(x)\,-\,\cos(x)\,-\,2\cdot\underbrace{\sin^2(x)}\cdot\cos(x)\)

\(\displaystyle \;\;\;=\;2\cdot\cos^3(x)\,-\,\cos(x)\,-\,2\cdot[\overbrace{1\,-\,\cos^2(x)}]\cdot\cos(x)\)

\(\displaystyle \;\;\;=\;2\cdot\cos^3(x)\,-\,\cos(x)\,-\,2\cdot\cos(x)\,+\,2\cdot\cos^3(x)\)

Therefore: \(\displaystyle \:\cos(3x)\;=\;4\cos^3(x)\,-\,3\cdot\cos(x)\)


The problem becomes:
\(\displaystyle \;\;\cos(3x)\,+\,\cos(x) \:=\:[4\cdot\cos^3(x)\,-\,3\cdot\cos(x)]\,+\,\cos(x)\:=\:4\cdot\cos^3(x)\, -\, 2\cdot\cos(x)\)