
Subject: PreCal ques: trig
I have some PreCal questions. I am a student at the secondary level. I
would be very grateful for your help.
Solve the equation for theta (0 <= theta < 2pi).
 tan^{2}(theta) = 3
I know sec^{2}(theta) 1 = tan^{2}(theta)

sin(theta) = cos(theta)
How would I go about solving these equations?
also:
In traveling across flat land, you notice a mountain directly in front
of you. Its angle of elevation (to the peak) is 3.5 degrees. After you
drive 13 miles closer to the mountain, the angle of elevation is 9 degrees.
Approximate the height of the mountain.
I solved it this way, but the answer doesn't seem right.
(sin5.5 degrees)/(13) = (sin3.5 degrees)/(b) ; b is the side
opposite the angle 3.5
8.28 mi = b
(sin90 degrees)/(8.28) = (sin9 degrees)/(x) ; x is the height of the
mountain
1.3 mi = x
The model for the height "h" of a Ferris wheel car is
h = 51 + 50sin[(8pi)t]
where "t" is measured in minutes. (The Ferris wheel has a radius of 50 ft.)
This model yields a height of 51 feet when t = 0 . Alter the model so that
the height of the car is 1ft when t = 0.
I don't know what to do for this. I'm lost.
Thank you for your help,
Angela
Hi again Angela,
I am a very strong believer in drawing diagrams. In a math problem, if it makes any sense at all to do so, draw a diagram.

From what you have said sec^{2}(theta) = 4. Also since (0 ≤ theta < 2pi), sec(theta) = 2. Now draw a right triangle with sec(theta) = 2. What is the length of the third side? do you recognize this triangle? What is theta? 


Again, draw a diagram. This time draw a right triangle and label its sides a, b and the hypotenuse c and label one angle as theta. Write sin(theta) and cos(theta) in terms of a, b and c. Set these expressions equal. What does this say about a and b? What is theta?

Again a diagram is appropriate. Here h is the height of the mountain in miles. From the diagram tan(3.5^{o}) = h/(13 + x) and tan(9^{o}) = h/x. Solve the second equation for x, substitute in the first equation and solve for h.



Here I don't see that a diagram helps. I think of two people building a model for the height of the Ferris Wheel. One of them uses the letter t to designate time and starts his clock when the height is 51 feet. He gets the model
h = 51 + 50sin[(8pi)t]
The second person uses the letter t^{*} to designate time and starts her clock when the height is 1 foot. In the first model h = 1 foot when sin[(8pi)t] = 1, that is when (8pi)t = (3/2)pi or t = 3/16 minutes. Hence the first person's clock is always 3/16 minutes ahead of the second person's clock, that is t = t^{*} + 3/16. The second person then just has to substitute this value for t into the first model to et her model.
Cheers,
Harley
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