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Homework Help: Solving for theta

  1. Sep 28, 2011 #1
    1. The problem statement, all variables and given/known data
    i dont think this should be too hard, i just forget how to do it...
    so i have this problem, and i've got it down to this

    -490/sin2[itex]\theta[/itex] + 2500cos[itex]\theta[/itex]/sin[itex]\theta[/itex] - 1800 = 0

    if anyone can help that would be great i just forget how to do this i tried using trig identites but i couldnt figure it out
  2. jcsd
  3. Sep 28, 2011 #2
    First you might try putting the entire thing as a fraction, then setting the numerator to zero (it should be a quadratic equation in sin2.
  4. Sep 28, 2011 #3
    -1800sin2[itex]\theta[/itex] + 2500cos[itex]\theta[/itex]sin[itex]\theta[/itex] -490 = 0

    i know how to do that, i just don't know from here how to solve for [itex]\theta[/itex]
  5. Sep 28, 2011 #4
    You have an equation of the form ax2+bx+c = 0. Can you find the roots?
  6. Sep 28, 2011 #5
    x=0.236, 1.153

    sorry not so good at writing equations in html...

    can you do that with the cos[itex]\theta[/itex] in there though? i would understand if it was
    asin2[itex]\theta[/itex] + bsin[itex]\theta[/itex] + c

    which would leave you with

    sin[itex]\theta[/itex]= root 1, root 2

    but i don't understand what to do with the roots from the top equation
  7. Sep 28, 2011 #6
    Offhand, I don't see how to get an explicit function for theta either. If you are only looking for a number that satisfies the equation, you can solve it numerically without using calculus or a computer.

    Sometimes, in industry, you wind up with transcendental equations and a numerical approach is the only way to solve them.
  8. Sep 28, 2011 #7


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    1/sin2θ = csc2θ

    csc2θ = 1 + cot2θ

    cosθ/sinθ = cotθ

    Plug those into the original & get a quadratic in cotθ .
  9. Sep 28, 2011 #8
    Yeah, now that i see the numbers, it looks like it is only solvable numerically. Are you sure the problem is written correctly (or that you didn't make a calculation mistake arriving at the initial equation you posted?)

    Ot you can do what SammyS says....that's why he has the PhD (knows lots of tricks)
  10. Sep 28, 2011 #9


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    Well, it's a messy little problem and you are right about using trig identities (you just don't use the usual ones).

    Replace [itex]\sin^{2} \theta [/itex] with [itex]\frac{1}{2} ( 1 - \cos 2\theta) [/itex] and [itex] \sin \theta \cos \theta [/itex] with [itex]\frac{1}{2} \sin 2\theta [/itex] to obtain*

    [tex] 900 ( 1 - \cos 2\theta) - 1250 \sin 2\theta = -490 . [/tex]

    * This first identity comes from manipulating the "cosine-double-angle formula" and the Pythagorean Identity

    Multiply the first term out and rearrange this to put the "double-angle" terms on one side:

    [tex] 900 - 900 \cos 2\theta - 1250 \sin 2\theta = -490 \Rightarrow 900 \cos 2\theta + 1250 \sin 2\theta = 1390 [/tex]

    You could now temporarily designate [itex] \alpha = 2\theta [/itex] , which now leaves you with a linear equation in sin(alpha) and cos(alpha), which can be solved by using the "angle-addition formulas" and an auxiliary phase angle (this is a standard method). It will work because [itex]\sqrt{900^{2} + 1250^{2}} [/itex] > 1390 . This will give you two solutions for alpha ; keep in mind that this is twice theta, so there could be up to four in the principal circle, 0 ≤ theta < 2(pi).
    Last edited: Sep 28, 2011
  11. Sep 28, 2011 #10

    Thanks a lot! i used this method and got 39.9o and 14.4o
    double checked and these angles make sense in the original problem.
    thanks everybody else for contributions

    bit of a hard problem for first year intro physics huh...

    if anybody is interested the original problem was

    An enemy ship is on the east side of a mountain island. The enemy ship has manoeuvred to within 2500m of the 1800m-high mountain peak and can shoot projectiles with an initial speed of 250m/s. If the western shoreline is horizontally 300m from the peak, what are the distances from the western shore at which a ship can be safe from the bombardment of the enemy ship?
  12. Sep 28, 2011 #11


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    This is where knowing the application saves a little work. The method I described reduces the problem to solving [itex] \sin (2\theta + 35.76º ) \approx .9024 [/itex], which gives four solutions between 0º and 360º . Two of them are the solutions you found and the other two are the angles with identical tangent values, that is, the angles 180º away. Your problem only needs (presumably) the angles in the first quadrant, so you would stop once you have those two.
  13. Oct 1, 2011 #12


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    An alternative way, which is not all that alternative:

    If A is the angle of firing, for horiz motion, you derived 1800=250.t.cosA
    so, rearranging, obtain cos A =72/t

    Draw a right angle triangle, mark one acute angle A, and appropriately two sides "72" and "t" in order that cos A=72/t
    Pythagoras gives you the other side.
    Using that triangle, you can see that sin A = sqrt(t2-722)

    Substitute into your vertical motion equation for sinA and cosA and you end up with a quadratic in t2. Solve for t2.
    Of the two positive values for t, any less than 10 secs can be discarded, since at 250 m/sec the shell will take at least 10 secs to cover the 2500+ distance.

    This leaves me with just one value, t=12.78secs.

    Knowing t, you can evaluate 72/t to find cosA.
    I get 38.5 degrees to just clip the mountain peak. It seems a bit low, but I haven't checked my working. We customarily anticipate there would be two firing angles where the shell will clip the peak, but it seems here that they can't pack enough cordite for that. :smile:

    So instead of needing to remember trig identities, you can use Pythagoras.
    Other than that, there is no difference in methods.
    Last edited: Oct 1, 2011
  14. Oct 1, 2011 #13


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    14.4o :confused: Any firing lower than 35.75o will just be landscaping.
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