Pre-Calculus Final t/m, can anyone help?

Pre-College Issues High School Life
1

<p>I have this 100 question review that has to be completed, or there is an automatic 10 pt deduction from your final grade . . and I haven't done this stuff since Jan, so if anyone could help . . </p>

<p>A plane travels 120 miles at a heading of N 25 W. It then changes direction and travels 135 miles at a heading of N 47 W. How far is the plane from its original position?
A) 225.3 mi
B) 250.3 mi
C)237.8 mi
D) 255 mi</p>

<p>Find all solutions of the equation in the interval [0,2pi)</p>

<p>tan^2x = sqrt2/2 * sec x</p>

<p>A) pi/4 , 7pi/4
B) 5pi/6 , 7pi/6
C) 3pi/4, 5pi/4
D) None of these</p>

<p>Thank you soo much in advance if anyone knows how to solve these 2 particular problems. Test tomorrow, 1st period =&lt;/p>

2

<p>so how did it go?</p>

3

<p>it's tomorrow, I have it in about 15 hours.
I'd use my calc for the 2nd one, but I don't think the calculator takes any of the functions ^2
And I'm not even sure how to draw the 1st one..</p>

4

<p>1 is 255
2 is unreadable. retype problem with parentheses and other helpful notations</p>

<p>i can help you on aim (hmm)</p>

5

<p>but could you tell me how you got 255?</p>

<p>noo, i can't get on aim yet, i'll get too distracted</p>

<p>tan^2x = (sqrt2/2) (sec x)
so . . do we start by converting tan^2x to sec^2x - 1?
I did that, and it got kinda messy</p>

6

<p>ok, first of all, the correct notation is (tan(x))^2</p>

7

<p>no, that reads as x^2 on the calculator, not tan^2</p>

8

<p>on #2, take the answers and plug them in the equation. you don't need tangent squared, just do tan x and square it</p>

9

<p>nevermind, 2 is A. you get it by simply typing ((tan(x))^2)/sec(x)-(sqrt2)/2 into the graphing calculator, and then find the zeroes in the graph mode</p>

10

<p>Alright, but surely there's a way to solve it by hand? There's kind of a time limit, plugging it in will take a while, and I tried that spect, and my calc says busy for a really long time. I guess I'll just find the teacher t/m morning</p>

<p>But could you tell me how you got 255 please? You can't just add the 2 distances, there's some trig involved</p>

11

<p>Sarorah:</p>

<ol>
<li>I don't think the answer is 255 miles. If the plane had continued on its original heading of N 25 W after the first 120 miles, it would be exactly 120+135 = 255 miles from its original position. Since it deviated from its original course, it will be < 255 miles from the start.</li>
</ol>

<p>Let A be the starting point, and let B be the ending point after the first leg. Extend AB arbitratrily beyond B, and draw BC at an angle of (47-25) = 22 degrees away from AB<em>extended, so that BC (the second leg of the trip) = 135. Now drop a line from C to AB</em>extended, so that it ends at a point D on AB_extended, and angle CDB is 90 degrees. BCD should now be a right-angled triangle, with angle B = 22 degrees, and BC = 135 .</p>

<p>Now, BD = BC cos(B) = 135 cos(22 degrees)
and CD = BC sin(B) = 135 sin( 22 degrees)</p>

<p>Draw the line AC. This is the hypotenuse of a larger right-angled triangle ACD, and length(AC) = distance of C from the starting point A.</p>

<p>AC^2 = CD^2 + AD^2
so AC = sqrt( CD^2 + (AB + BD)^2 )
= sqrt( (135 sin(22))^2 + (120 + 135 cos(22))^2 )</p>

<p>Plug in the numbers for sin(22) and cos(22), and see what you get.</p>

<ol>
<li>If I understand your question correctly, you need to solve for x so that
[tan(x)]^2 = sqrt(2) sec(x) /2
or [sin(x) / cos(x)]^2 = 1/ (sqrt(2) .cos(x))
[sin(x)]^2 = cos(x) / sqrt(2)
1 - [cos(x)]^2 = cos(x) / sqrt(2)</li>
</ol>

<p>For simplicity, set y = cos(x). Then the equation becomes
1 - y^2 = y / sqrt(2)
sqrt(2) - sqrt(2).y^2 - y = 0
sqrt(2).y^2 + y - sqrt(2) = 0</p>

<p>This is a quadratic in y, solve:
y = [ -1 +/- sqrt( 1 + 4 sqrt(2)sqrt(2)) ] / [2 sqrt(2)]
= -1 +/- sqrt(9) / [2 sqrt(2)]
= 2 / 2 sqrt(2) or -4 / 2 sqrt(2)
= 1 / sqrt(2) or - sqrt(2) ; discard the latter, it's > 1 and infeasible for cos(x)</p>

<p>So y = cos(x) = 1/sqrt(2) , implying that x = 45 degrees or 315 degrees
In radians, that's pi/4 or 7 pi/4 .</p>

12

<p>sarah:</p>

<p>for number 1, use the law of cosine</p>

<p>A^2 = B^2 + C^2 - 2BC*cos(a)</p>

<p>In this case, B=135, C=120, and a(the angle)= 22 (i'm assuming this is radian because otherwise none of the choices work.</p>

13

<p>Spectrum303's method is way faster than what I had proposed. Strictly speaking, the angle a should be 180 - 22 = 158, I think.</p>

14

<p>i think going from N25W to N47W is only a 22 radian change</p>

15

<p>I think your first method works too, also I tried drawing it as an obtuse triangle, and got the 2 sides of 120 and 135 with an included angle of 158 degrees, and using the law of cosines, gives you a third side of 250.3, letter B.</p>

<h1>2, I think your answer is right, but when you do</h1>

<p>[tan(x)]^2 = sqrt(2) sec(x) /2
or [sin(x) / cos(x)]^2 = 1/ (sqrt(2) .cos(x))</p>

<p>what happened to the 2 that the right side was originally divided by?</p>

<p>And one more question if you don't mind, this one:
Verify the identity
cos (x - 3pi/2) = - sin x
I converted to degrees and got
cos (x - 270) = - sin x</p>

<p>Well, the cos (90 - x) = sin x
and cos (180) = -1</p>

<p>But, cos (90-x) is different from cos (x-90), right?
Is there another way to verify the identity?
Thanks again for the help, really appreciated.</p>

16

<p>Yep, I did it the 2nd method, I drew these parallel lines and you get a 22 degree change, which subtracted from 180 = 22</p>

17

<p>once again, i can be much more efficient on aim, but whatever</p>

<p>cos is an even function. this means that cos(90-x) is the exact same thing as cos(x-90)</p>