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Math Help - Trigonometry

  1. #1
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    Trigonometry

    Hey, I have a whole sheet of trigonometry questions to do and I've drawn just 4 of them in paint. Please take me through them as I haven't a clue. Many thanks!

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  2. #2
    TD!
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    In a right triangle, the sine of one of the (non-right) angles is equal to the opposite side divided by the hypotenuse while the cosine of that angle is the adjacent side divided by the hypotenuse.

    For problem 1, this gives that cos(27) = x/5.89 which is an equation you can solve for x. The other problems are similar!
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  3. #3
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    So my answer for 1 would be x= 5.2480 (4dp)

    Am I right?

    Can you give me the equation used, as you did for cos(27) = x/5.89, for the others please?
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  4. #4
    TD!
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    Quote Originally Posted by tristan1
    So my answer for 1 would be x= 5.2480 (4dp)
    Am I right?
    That's correct

    Quote Originally Posted by tristan1
    Can you give me the equation used, as you did for cos(27) = x/5.89, for the others please?
    Well I could, but then you wouldn't be learning anything from it.
    Try to set-up such an equation youself. I'll list the possible formula's you can use:

    \begin{gathered}<br />
  \sin \alpha  = \frac{{{\text{opposite}}}}<br />
{{{\text{hypotenuse}}}} \hfill \\<br />
  \cos \alpha  = \frac{{{\text{adjacent}}}}<br />
{{{\text{hypotenuse}}}} \hfill \\<br />
  \tan \alpha  = \frac{{{\text{opposite}}}}<br />
{{{\text{adjacent}}}} \hfill \\ <br />
\end{gathered}

    Now: first identify which of the three is given (the opposite, adjacent or hypotenuse) and which is asked.
    Then choose the formula which has both these sides and try to set up the equation.
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  5. #5
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    OK then, can you explain why I used COS for question one.
    I really dont get how you determin which equation you use.
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  6. #6
    TD!
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    Of course, as I said: you have to determine which sides are involved in the problem.

    Do you understand the terms adjacent, opposite and hypotenuse?
    Here's a clear image, be careful because it's important which angle is marked.



    Now, for the marked angle at the right, the bottom side is 'adjacent' (it touches the angle but isn't the longest side), the left side is 'opposite' (it doesn't touch the angle) and the longest side is called the hypotenuse.

    In the first problem, the marked angle is at the top left. The side we're looking for is the top side, which is the adjacent side for this angle. The hypotenuse is given, so we're looking for the formula which involves the hypotenuse and the adjacent side, that's the on with the cosine.

    Do you understand?
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  7. #7
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    OK, I kinda understand. For 2 (and 3), I need to find the hypotonuse.

    So I'm looking for one of those 3 equations to contain the ajacent and hypotenuse. Therefore I'd use COS?

    For 2, I'd type in COS 55 in the calculator then times the answer by 45?

    For 3, I'd type COS 71 in the calculator the tiems by 163. Are these correct?
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  8. #8
    TD!
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    Take them one at a time and look carefully.

    For two, the marked angle is at the top left. The hypotenuse is indeed the side we're looking for, but which one is given (from the point of view of the angle): the adjacent or opposite side?
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  9. #9
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    Opposite?
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  10. #10
    TD!
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    Indeed! The side which isn't linked to the angle, that is the opposite side.
    So what formula will you be using? You need one that involed the hypotenuse and the opposite side.
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  11. #11
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    SIN?

    So how do I approach this in a calculator?
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  12. #12
    TD!
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    Indeed sin. Now write out the formula I gave for the sine, replacing all expressions you know with their values and call the unknown side x.
    Then solve for x, the expression you now have can be entered in a calculator.
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  13. #13
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    SIN 55 =0.8191

    0.8191 x 45 = 36.8618

    Right?
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  14. #14
    TD!
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    You should first try to set-up the equation correctly, starting from the formula.
    Write down the formula of the sine and fill in eveything you know, i.e. replace "opposite", "hypotenuse" and "alpha" with their values (or x, if it's the unknown).

    \sin \alpha  = \frac{{{\text{opposite}}}}<br />
{{{\text{hypotenuse}}}} \Rightarrow \sin 55^\circ  = \frac{{45}}<br />
{x}

    You then solve for x

    \sin 55^\circ  = \frac{{45}}<br />
{x} \Leftrightarrow x\sin 55^\circ  = 45 \Leftrightarrow x = \frac{{45}}<br />
{{\sin 55^\circ }} \approx 54.93
    Last edited by TD!; January 13th 2006 at 11:16 AM.
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