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    Ok, so I was trying to do some work to prepare for A levels, but I keep getting the wrong answer to a certain question. I can't work out exactly where I have gone wrong, but I keep getting the answer of 8.01m when the answer sheet says that the answer is 7.52m. Could someone possibly give me a step by step method so that I can see exactly where I keep going wrong?

    The question is to find the value of x and this is the diagram: http://tinypic.com/view.php?pic=bi19w7&s=9#.WXjmw73TXqA
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    (Original post by Coco_Monkey)
    Ok, so I was trying to do some work to prepare for A levels, but I keep getting the wrong answer to a certain question. I can't work out exactly where I have gone wrong, but I keep getting the answer of 8.01m when the answer sheet says that the answer is 7.52m. Could someone possibly give me a step by step method so that I can see exactly where I keep going wrong?

    The question is to find the value of x and this is the diagram: http://tinypic.com/view.php?pic=bi19w7&s=9#.WXjmw73TXqA
    HOLD THE FOOK UP, are you doing a level work in the holidays?
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    (Original post by sweet boner)
    HOLD THE FOOK UP, are you doing a level work in the holidays?
    Yes, one of the sixth forms that I applied to wants us to prepare for a test that there will be on the first week. So I'm doing some stuff in case I end up going there.
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    (Original post by Coco_Monkey)
    Yes, one of the sixth forms that I applied to wants us to prepare for a test that there will be on the first week. So I'm doing some stuff in case I end up going there.
    Ooooh good luck with the integration, trigonometric equations, Pythagorean identities and all that *******s
    you get the idea
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    (Original post by sweet boner)
    Ooooh good luck with the integration, trigonometric equations, Pythagorean identities and all that *******s
    you get the idea
    Can't you help me?! And just looking at the names of those makes me want to cry a little on the inside
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    Can't you just use sine/cosine rule?
    Anyway, what do you think you got the maths gcse? I got around 125, is that a 6?
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    (Original post by Coco_Monkey)
    Can't you help me?! And just looking at the names of those makes me want to cry a little on the inside
    Just by looking at it, i say use pythag to find the hypotenuse.
    You know you have 1 right angle and 70 degree angle soo.....
    180- (70-90)= You get another angle then it's just easy work.
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    (Original post by Vexillarius)
    Can't you just use sine/cosine rule?
    Anyway, what do you think you got the maths gcse? I got around 125, is that a 6?
    It's in the trigonometry section, so I'm guessing it wants me to use that. And I have no clue about grade boundaries
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    (Original post by Coco_Monkey)
    It's in the trigonometry section, so I'm guessing it wants me to use that. And I have no clue about grade boundaries
    Btw, sine and cosine rule are both uses of trigonometry AFAIK.
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    (Original post by Vexillarius)
    Btw, sine and cosine rule are both uses of trigonometry AFAIK.
    I meant the whole sin, cos, tan area because that is the only stuff mentioned in the lesson part of the sheet.
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    (Original post by Coco_Monkey)
    It's in the trigonometry section, so I'm guessing it wants me to use that. And I have no clue about grade boundaries
    Calculate the hypotenuse (y) of the 60 degree triangle. That is the opposite for the 70 degree triangle.

    cos(60)=\frac{11}{y}

    \therefore y=\frac{11}{cos(60)}

    For the second triangle, you have the opposite (y) and the adjacent (x). Using tan(70) you should be able to calculate x.
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    (Original post by Coco_Monkey)
    I meant the whole sin, cos, tan area because that is the only stuff mentioned in the lesson part of the sheet.
    In the left/up triangle, you can find the length of its hypotenuse by noting that \cos 60^{\circ} = \frac{a}{h} where you know the adjacent a = 11. So you can get h = \frac{a}{\cos 60^{\circ}}. Now that you have the hypoteneuse of the first triangle, you can use \tan 70^{\circ} = \frac{o}{x} where the opposite of 70 degrees is the hypoteneuse of the first triangle (whose length you've found).

    You can now use this to rearrange for x.
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    (Original post by RogerOxon)
    Calculate the hypotenuse (y) of the 60 degree triangle. That is the opposite for the 70 degree triangle.

    cos(60)=\frac{y}{11}

    \therefore y=11cos(60)

    For the second triangle, you have the opposite (y) and the adjacent (x). Using tan(70) you should be able to calculate x.
    cos 60 would be adjacent/hyp, so 11/y, surely?
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    So is the hyp of the first triangle (the diagonal between the two triangles) 22 then? Because that's what I got originally before asking this question.
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    (Original post by AnIndianGuy)
    Yes Roger got it wrong.

    You could use sin 30 too obvs
    Oops. Yes.
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    (Original post by rogeroxon)
    oops. Yes.
    roger help me
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    (Original post by Coco_Monkey)
    So is the hyp of the first triangle (the diagonal between the two triangles) 22 then? Because that's what I got originally before asking this question.
    yes
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    (Original post by Coco_Monkey)
    roger help me
    The diagonal is y:

    y=\frac{11}{cos(60)}

    Substitute for cos(60) to simplify y.

    tan(70)=\frac{y}{x}

    Rearrange and you'll get the answer.
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    (Original post by Zacken)
    yes
    Then how does x = 7.52? Have I used the wrong ratio to find the answer? I used Tan.
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    (Original post by Coco_Monkey)
    Then how does x = 7.52? Have I used the wrong ratio to find the answer? I used Tan.
    how did you use tan
 
 
 
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