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Math Help - Trig / right angled triangles / SIN COS TAN

  1. #1
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    Question Trig / right angled triangles / SIN COS TAN

    Hi all,
    My first post here so my apologies in advance if this is not the correct place to post this.

    I have a right angled triangle with one angle of 25˚ (so clearly the last is 65˚). The hypotenuse is 17cm.

    My question is how do I work out the other two sides please?

    I'm guessing I have to use SIN and have found that:

    SIN(A) = opposite(o)/hypotenuse(h)
    SIN(65˚) = o/17
    SIN(65˚) x 17 = o
    0.8268 x 17 = o
    o = 14.05

    Am I on the right track?

    Any help greatly appreciated.
    Thank you in advance.
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  2. #2
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    Quote Originally Posted by cyberdyne View Post
    Hi all,
    My first post here so my apologies in advance if this is not the correct place to post this.

    I have a right angled triangle with one angle of 25˚ (so clearly the last is 65˚). The hypotenuse is 17cm.

    My question is how do I work out the other two sides please?

    I'm guessing I have to use SIN and have found that:

    SIN(A) = opposite(o)/hypotenuse(h)
    SIN(65˚) = o/17
    SIN(65˚) x 17 = o
    0.8268 x 17 = o
    o = 14.05

    Am I on the right track?

    Any help greatly appreciated.
    Thank you in advance.
    Looks OK to me, especially since you're not given which angle is 25 degrees.

    To find the last side use Pythagoras' Theorem

    edit: What masters' said, especially since I didn't check the arithmetic
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  3. #3
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    Quote Originally Posted by cyberdyne View Post
    Hi all,
    My first post here so my apologies in advance if this is not the correct place to post this.

    I have a right angled triangle with one angle of 25˚ (so clearly the last is 65˚). The hypotenuse is 17cm.

    My question is how do I work out the other two sides please?

    I'm guessing I have to use SIN and have found that:

    SIN(A) = opposite(o)/hypotenuse(h)
    SIN(65˚) = o/17
    SIN(65˚) x 17 = o
    0.8268 x 17 = o
    o = 14.05

    Am I on the right track?

    Any help greatly appreciated.
    Thank you in advance.
    Hi cyberdyne,

    Yes, you're on the right track except sin (65) = .9063.
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  4. #4
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    Quote Originally Posted by cyberdyne View Post
    Hi all,
    My first post here so my apologies in advance if this is not the correct place to post this.

    I have a right angled triangle with one angle of 25˚ (so clearly the last is 65˚). The hypotenuse is 17cm.

    My question is how do I work out the other two sides please?

    I'm guessing I have to use SIN and have found that:

    SIN(A) = opposite(o)/hypotenuse(h)
    SIN(65˚) = o/17
    SIN(65˚) x 17 = o
    0.8268 x 17 = o
    o = 14.05

    Am I on the right track?

    Any help greatly appreciated.
    Thank you in advance.
    Hi cyberdyne,

    welcome to MHF.

    sin55.77^o=0.8267

    sin65^o=0.9063

    sin25^o=0.422

    Then you can calculate the lengths of the sides opposite those angles as you were doing

    Of course you could also use cos65^o instead of sin25^o for the third side

    sin65^o=\frac{opp}{hyp}=\frac{opp}{17}

    17sin65^o=opp

    calculates the length of the side opposite the 65 degree angle.
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  5. #5
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    Quote Originally Posted by e^(i*pi) View Post
    Looks OK to me, especially since you're not given which angle is 25 degrees.

    To find the last side use Pythagoras' Theorem
    Wow, thanks for the quick reply.

    So;
    a = 17 - o
    a = 289 - 197.4025
    a = 91.5975
    a = sq rt of 91.5975
    a = 9.57065828
    a = 9.57

    (it's been a long time since I did these)
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    Hi cyberdyne,

    welcome to MHF.

    sin65^o=0.9063
    Quote Originally Posted by masters View Post
    Hi cyberdyne,

    Yes, you're on the right track except sin (65) = .9063.
    Ah, OK, thank you. Not sure where I got my calculation from.
    And thank you for the welcome.
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  7. #7
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    OK so I have:

    c=17

    SIN(A) = opposite(b)/hypotenuse(h)
    SIN(65˚) = b/17
    b = SIN(65˚) x 17
    b = .9063 x 17
    b = 15.4071

    c = a + b
    17 = a + 15.4071
    a = 17 - 15.4071
    a = 289 - 237.37873
    a = 51.62127
    a = sqrt 51.62127
    a = 7.184

    so a=7.184, b=15.4071, c=17

    7.184 + 15.4071 = 288.988586

    Presumably you cannot always obtain three exact figures?
    Thanks again.
    Last edited by cyberdyne; May 7th 2010 at 02:25 PM.
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  8. #8
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    Yes,

    you have some "round-off", if you work with a number of significant figures.

    17sin65^o=17cos25^o=7.18451044959

    17sin25=17cos65^o=15.4072323796

    These are the lengths of the perpendicular sides.

    (7.18451044959)^2+(15.4072323796)^2=289=17^2
    Last edited by Archie Meade; May 8th 2010 at 03:41 AM.
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  9. #9
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    I can relax now!
    Many thanks for your help.
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  10. #10
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    A couple of extra questions if I may. My daughter had this question in a mock exam paper and I'm trying my best to help her, having not studied Trig myself for approx. 24 yrs!

    1. Can anyone please explain how I would find the SIN(E) of 65 degrees on a scientifc calculator. My error above was from entering sin(65) into Google [0.826828679] instead of sin(65 degrees) [0.906307787] but I cannot seem to get the correct value on a calc. It is a relatively basic Texet FX1000
    2. Does this question seem a little advanced for a 14 yr old to anyone else? I can't recall using COS, SIN & TAN until my 5th yr of secondary school (todays' yr 11) but my daughter is currently in yr 9.

    Many thanks again.

    EDIT:
    Got it. For some unknown reason I had to hit the RESET button on my calc.
    Thanks
    Last edited by cyberdyne; May 8th 2010 at 04:52 AM.
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  11. #11
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    Quote Originally Posted by cyberdyne View Post
    A couple of extra questions if I may. My daughter had this question in a mock exam paper and I'm trying my best to help her, having not studied Trig myself for approx. 24 yrs!

    1. Can anyone please explain how I would find the SIN(E) of 65 degrees on a scientifc calculator. My error above was from entering sin(65) into Google [0.826828679] instead of sin(65 degrees) [0.906307787] but I cannot seem to get the correct value on a calc. It is a relatively basic Texet FX1000
    2. Does this question seem a little advanced for a 14 yr old to anyone else? I can't recall using COS, SIN & TAN until my 5th yr of secondary school (todays' yr 11) but my daughter is currently in yr 9.

    Many thanks again.

    EDIT:
    Got it. For some unknown reason I had to hit the RESET button on my calc.
    Thanks
    Hi cyberdyne,

    when you calculated sin65

    the calculator was in radian mode.
    The online calculator needed to be in degree mode,
    so just set your calculator to degree mode using "deg".
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  12. #12
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    Quote Originally Posted by Archie Meade View Post
    Hi cyberdyne,

    when you calculated sin65

    the calculator was in radian mode.
    The online calculator needed to be in degree mode,
    so just set your calculator to degree mode using "deg".
    Hi Archie,
    This is what I realised after reading the same elsewhere, but no matter what I pressed I could not obtain the correct firgure. I have a feeling the calculator was stuck in radian mode for some reason, hence the RESET. It worked though so all is well.
    Many thanks
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