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Think Surdic question

  1. Jul 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Hey,

    Right my problem is this. I have not been doing maths for over a year and a bit. And I am very rusty.

    So I am stuck on what this question is meaning, and what to do.

    Find the sin, cos, and tan of 60', 30' and 45'. [' is degree.]
    Leave your answer in surdic form. [Root 2 is called a surd.]

    2. Relevant equations
    The words in bold are the question from my paper I have been sent.
    Now this is what I am stuck on. What in heck are they on about? And what are they meaning.

    3. The attempt at a solution
    I tried it and thought they were meaning that the values for the sin cos tan, were root3/2, root3/2, 1. But I am far from sure if it the answer or not.


    Cheers,

    venito
     
  2. jcsd
  3. Jul 20, 2009 #2

    berkeman

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    Staff: Mentor

    Thread moved from calc to pre-calc math forum.

    Welcome to the PF. To put the answers in a form that has the square roots in it, remember the Pythagorean theorem and draw triangles with those degree angles in them. For example, the right triangle with 30' and 60' angles in it has side lengths of what?
     
  4. Jul 20, 2009 #3
    Usually, trig students memorize these values. Remember to not have a square root in the denominator!
     
  5. Jul 20, 2009 #4
    Thanks did not know were to place it.

    Thanks mate,

    Right so I am trying it now. So it is a triangle with A C B. A is on the left hand side, and is the tip of the triangle. C and B are the bottom of it, which is too the right.

    Length of A-C is 4.8. C-B is 3.9.

    So now I guess I do this. I have drawn a triangle, and added the values of 30' and 60 to it. Which has lengths of 4.8 and 2.9. And will do another triangle for the 45' angle. And I guess with the 45' and angel has the same lengths?

    Right so far I just have a triangle, with numbers. Is there a formula or something?
     
  6. Jul 20, 2009 #5

    berkeman

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    Try drawing your triangles with unit values for the sides. Like, the simplest triangle with 90', 45' and 45' angles, has sides of 1, 1, ____. And the simplest right triangle with angles of 30' and 60' has sides of length 1, ____, ____.
     
  7. Jul 20, 2009 #6
    Right so to answer this.

    Okay now I have the triangles like that. So do I now plug in the other values I have or do they get some sort of multiplication?

    mac
     
  8. Jul 20, 2009 #7

    berkeman

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    Just use the definitions of the sin, cos and tan functions, and those triangles to fill out the table of answers. And remember Pinu7's comment about getting rid of any roots in denominators (those are usually not considered simplified).
     
  9. Jul 20, 2009 #8
    Right. I will let you know how I go.

    Thanks.
     
  10. Jul 20, 2009 #9
    Right so I played with it.

    And have come up with this. 1/2, 0.707106781, √3/2. √3/2, 0.707106781, 1/2. 0.577350269, 1, √3.

    Which is the memorized functions or answers to sin cos tan 30 45 60. But with no square roots on the denominator.

    So in other words they just want all 9 forms of the above numbers?
     
  11. Jul 20, 2009 #10

    berkeman

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    They want a 3x3 table with the answers... put 30' 45' 60' across the top as colunm headings, and sin, cos, tan as row labels. Fill in each of the 9 boxes with an answer that does not have a decimal point in it.
     
  12. Jul 20, 2009 #11
    Right so that means I am correct. So what is Pinu7 meaning then about the roots in the denominator?

    Should they be taken out or not, from the box of nine I have, or what I mean the results from sin 30 45 60 cos 30 45 60 tan 30 45 60.
     
  13. Jul 20, 2009 #12

    berkeman

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    I didn't work out the table, so I don't know offhand if you will run into the situation or not. But generally in algebra, when they ask you to simplify an expression, and you end up something like:

    [tex]\frac{1}{\sqrt{2}}[/tex]

    then you need to get rid of the radical in the denominator to get full credit for the simplification. What could you multiply both the numerator and denominator by, in order to get rid of the square root in the denominator?
     
  14. Jul 20, 2009 #13
    So I would then take the problem.

    [tex]\frac{1}{\sqrt{2}}[/tex]

    Then multiply top and bottom by √2/√2. Which would then give me a √2.
    Is that right?
     
  15. Jul 20, 2009 #14

    berkeman

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    Not quite. Be sure to multiply both top and bottom by √2...
     
  16. Jul 20, 2009 #15
    What a idiot. It is √2/2.

    Which is now correct.
     
  17. Jul 20, 2009 #16

    berkeman

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    Correct-amundo!
     
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