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Obstuse trignometry

  1. Jul 5, 2006 #1
    1) given that A is obstuse and that sinA=2/(sqaureroot of 5). find the value of cosA and tanA.
    2)given that tanA=-5/12 and that tanA and cosA have opposite signs, find the values of sinA and of cosA.

    CAN SOMEONE TEACH ME EXACTLY HOW TO DO THESE PROBLEMS WITHOUT USING TRGNOMETRIC IDENTITES.PLEASE....I AM NEW TO THIS TOPIC AND I WANT TO GET A HEADSTART OVER MY PEERS.
    THANKS A LOT
    VIJAY
     
  2. jcsd
  3. Jul 5, 2006 #2

    VietDao29

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    Homework Helper

    Ok, if A is obtuse, what sign will sin, cos, and tan of the angle A take (is it positive or negative?). You can take a look at the unit circle to see. Given sin A, one can manage finding cos A using Pythagorean Identity, i.e:
    sin2 A + cos2 A = 1.
    Having sin A, and cos A, can you find tan A? [tex]\tan A = \frac{\sin A}{\cos A}[/tex]
    I'll give you an example:
    A is an acute angle. Given that sin A = 1 / 5. Find cos A, and tan A
    Since A is an acite angle (i.e < 900), cos A must be positive. Using the Pythagorean Identity, we have:
    [tex]\sin ^ 2 A + \cos ^ 2 A = 1[/tex]
    [tex]\Leftrightarrow \cos ^ 2 A = 1 - \sin ^ 2 A = 1 - \frac{1}{25} = \frac{24}{25}[/tex]
    [tex]\Rightarrow \cos A = \pm \frac{\sqrt{24}}{5}[/tex]
    Since cos A is positive, we choose:
    [tex]\cos A = \frac{\sqrt{24}}{5}[/tex]
    We have:
    [tex]\tan A = \frac{\sin A}{\cos A} = \frac{1}{\sqrt{24}}[/tex]
    Can you go from here? :)
     
  4. Jul 5, 2006 #3
    Thanks A Lot For The Solution
     
  5. Jul 5, 2006 #4
    vietdao...but is there any other way of doin such problems withou using the trignometric identities...because this problem is printed before the worksheet on trignometric identities....?
     
  6. Jul 5, 2006 #5
    yes..thanks a lot...i can do it at last....but...does obtuse mean it is in the second quadrant?
     
  7. Jul 5, 2006 #6
    Hello there ,
    First thing in determining the signs , i suggest you a quadrant . Hence , there will be a first quadrant , second quandrant , third and fourth . Marking each quadrants , from 3 o'clock ( anticlockwise direction ) A , S , T , C (Where A=all postive , S=Only Sinus Postive , T=Only Tangent Positive , C=Only Cosinus Positive) .
    Next you can then determine which quadrant the question is referring to . You can do without using trigonometric identities =) Try drawing and labelling the coordinates of your triangle , you should be able to find the answer =) Beware of the + and - signs .
     
  8. Jul 5, 2006 #7
    ok...thanks a lot..
     
  9. Jul 5, 2006 #8

    HallsofIvy

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    Draw a picture. Since you know [itex]sin(x)= \frac{2}{\sqrt{5}}[/itex] and sine is "opposite over hypotenuse", draw a right triangle with one side 2 and hypotenuse [itex]\frac{1}{\sqrt{5}}[/itex]. You can use the Pythagorean theorem to find the third side and calculate all the other functions from that. Then use the fact that the angle is obtuse (and so in the second quadrant) to determine the sign.
     
  10. Jul 6, 2006 #9
    yea...i get it....there r two ways of doing it..thanska lot
     
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