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Math Help - Stuck, it's fairly basic, please help!

  1. #1
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    Stuck, it's fairly basic, please help!

    5\sqrt{13}=a\sqrt{3}^2 + 10^2 - 2\times{a\sqrt{3}}\times{10}\times{cos150}

    How would i work out a? how do you square a\sqrt3 and 5\sqrt13

    Can someone please show me in stages how to fully expand - 2\times{a\sqrt{3}}\times{10}\times{cos150} also as i'm not show how when surds and trigonometric functions are involved.

    Many thanks guys!

    F.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Flexible View Post
    (5\sqrt{13})^2=(a\sqrt{3})^2 + 10^2 - 2a\sqrt{3}(10)\cos150

    How would i work out a? how do you square a\sqrt3 and 5\sqrt13

    Can someone please show me in stages how to fully expand - 2\times{a\sqrt{3}}\times{10}\times{cos150} also as i'm not show how when surds and trigonometric functions are involved.

    Many thanks guys!

    F.
    I assumed that you were using the law of cosines, so I tweaked your notation a little bit.

    (a\sqrt{3})^2=a^2 \cdot (\sqrt{3})^2=a^2 \cdot 3 = 3a^2

    (5\sqrt{13})^2=5^2 \cdot (\sqrt{13})^2=25 \cdot 13 = 325

    Start here:

    325=3a^2+100-20a\sqrt{3}\left(-\frac{\sqrt{3}}{2}\right)
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    Thanks! i don't really know how to manipulate the surds? hence i cant expand and arrange to find a?
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by masters View Post
    I assumed that you were using the law of cosines, so I tweaked your notation a little bit.

    (a\sqrt{3})^2=a^2 \cdot (\sqrt{3})^2=a^2 \cdot 3 = 3a^2

    (5\sqrt{13})^2=5^2 \cdot (\sqrt{13})^2=25 \cdot 13 = 325

    Start here:

    325=3a^2+100-20a\sqrt{3}\left(-\frac{\sqrt{3}}{2}\right)
    By the way, \cos 150=- \cos 30=-\frac{\sqrt{3}}{2}

    Now, to simplify:

    325=3a^2+100-20a\sqrt{3}\left(-\frac{\sqrt{3}}{2}\right)

    325=3a^2+100-10a\sqrt{3}(-\sqrt{3})

    325=3a^2+100+10a(3)

    325=3a^2+100+30a

    225=3a^2+30a

    3a^2+30a-225=0

    a^2+10a-75

    (a+15)(a-5)=0

    a+15 = 0 \ \ or \ \ a-5=0

    a=-15 \ \ or \ \ a=5

    a=-15 is obviously an extraneous root, so a = 5 is your solution.
    Last edited by masters; October 10th 2008 at 09:31 AM.
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  5. #5
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    Quote Originally Posted by Flexible View Post
    Thanks! i don't really know how to manipulate the surds? hence i cant expand and arrange to find a?
    The rule is \sqrt{ab}=\sqrt{a}\sqrt{b}
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    is there anywhere i can find a formulae sheet with the trigonometric functions in surd form?
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  7. #7
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    Quote Originally Posted by Flexible View Post
    is there anywhere i can find a formulae sheet with the trigonometric functions in surd form?
    You could start with the unit circle.
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  8. #8
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    Quote Originally Posted by Flexible View Post
    is there anywhere i can find a formulae sheet with the trigonometric functions in surd form?
    Yes, the Unit Circle helps, but it also helps to have the special triangles on hand so you can find the trig functions of the focus angles.

    Draw a right angled isosceles triangle with the identical sides = 1 unit in length. Clearly its angles are 45^\circ, 45^\circ, 90^\circ.

    By Pythagoras, the hypotenuse is \sqrt{2} units in length.

    What are \sin{45^\circ}, \cos{45^\circ} and \tan{45^{\circ}}?


    Then draw an equilateral triangle of side length 2 units. Draw a perpendicular bisector, then you have two 30^\circ, 60^\circ, 90^\circ triangles. Their side lengths are 1, \sqrt{3}, 2 units (check using Pythagoras).

    What are \sin{30^\circ}, \cos{30^\circ}, \tan{30^\circ}, \sin{60^\circ}, \cos{60^\circ} and \tan{60^\circ}?
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