Maths year 11

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    (Original post by B_9710)
    That's right.
    Thanks xxx

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    (Original post by z_o_e)
    Oh?
    Is this fine?

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    BCA is the angle you want to work out.

    You're given two sides, 12.8 - which is opposite to angle C and 6.8 which is adjacent to angle C

    You have opposite and adjacent sides which means you use Tan, more specifically since it's an angle you use the inverse of tan.

    So, arctan(12.8/6.8) = angle C



    arctan just means tan^-1 just easier to denote.
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    (Original post by 34908seikj)
    BCA is the angle you want to work out.

    You're given two sides, 12.8 - which is opposite to angle C and 6.8 which is adjacent to angle C

    You have opposite and adjacent sides which means you use Tan, more specifically since it's an angle you use the inverse of tan.

    So, arctan(12.8/6.8) = angle C



    arctan just means tan^-1 just easier to denote.
    Here xx



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    That's correct, did you see my post about part B?
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    (Original post by 34908seikj)
    That's correct, did you see my post about part B?
    Yes I did thanks ♥

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    (Original post by 34908seikj)
    BCA is the angle you want to work out.

    You're given two sides, 12.8 - which is opposite to angle C and 6.8 which is adjacent to angle C

    You have opposite and adjacent sides which means you use Tan, more specifically since it's an angle you use the inverse of tan.

    So, arctan(12.8/6.8) = angle C



    arctan just means tan^-1 just easier to denote.


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    Correcto.
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    (Original post by 34908seikj)
    Correcto.
    I didn't really understand this..

    So I'm suppose to find BD

    And what did X stand for ?
    And where did sin come from :/

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    (Original post by z_o_e)
    I didn't really understand this..

    So I'm suppose to find BD

    And what did X stand for ?
    And where did sin come from :/

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    Okay - x just represents the length of BD, it's arbitrary It could be anything.

    Do you understand right angled trigonometry? (Sin Opposite side/hypothenuse Cos adjacent side/hypothenuse tan opposite side/adjacent side)


    Now you're given an angle - 42 degrees and one side that is opposite from the angle which is 12.8 but we want to find the length of the hypothenuse

    so the two sides involved in the equation are the opposite and the hypothenuse, this means we use sin.


    so set it out: Sin(42) = 12.8 / the length of BD (in this case x)

    multiply both sides by x to get xsin(42) = 12.8 - the right hand side cancels out side you're also dividing by x


    Now to get x by itself divide both sides by sin(42) - since sin(42) is already being multiplied they will cancel out, leaving 12.8/sin(42)


    And that's it. x = 12.8/sin(42) = 19. something
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    (Original post by 34908seikj)
    Okay - x just represents the length of BD, it's arbitrary It could be anything.

    Do you understand right angled trigonometry? (Sin Opposite side/hypothenuse Cos adjacent side/hypothenuse tan opposite side/adjacent side)


    Now you're given an angle - 42 degrees and one side that is opposite from the angle which is 12.8 but we want to find the length of the hypothenuse

    so the two sides involved in the equation are the opposite and the hypothenuse, this means we use sin.


    so set it out: Sin(42) = 12.8 / the length of BD (in this case x)

    multiply both sides by x to get xsin(42) = 12.8 - the right hand side cancels out side you're also dividing by x


    Now to get x by itself divide both sides by sin(42) - since sin(42) is already being multiplied they will cancel out, leaving 12.8/sin(42)


    And that's it. x = 12.8/sin(42) = 19. something
    Got it ♥♥♥ and yes I love pythagoras and trigonometry so much!!!

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    (Original post by 34908seikj)
    Okay - x just represents the length of BD, it's arbitrary It could be anything.

    Do you understand right angled trigonometry? (Sin Opposite side/hypothenuse Cos adjacent side/hypothenuse tan opposite side/adjacent side)


    Now you're given an angle - 42 degrees and one side that is opposite from the angle which is 12.8 but we want to find the length of the hypothenuse

    so the two sides involved in the equation are the opposite and the hypothenuse, this means we use sin.


    so set it out: Sin(42) = 12.8 / the length of BD (in this case x)

    multiply both sides by x to get xsin(42) = 12.8 - the right hand side cancels out side you're also dividing by x


    Now to get x by itself divide both sides by sin(42) - since sin(42) is already being multiplied they will cancel out, leaving 12.8/sin(42)


    And that's it. x = 12.8/sin(42) = 19. something
    What about this question?


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    (Original post by 34908seikj)
    Okay - x just represents the length of BD, it's arbitrary It could be anything.

    Do you understand right angled trigonometry? (Sin Opposite side/hypothenuse Cos adjacent side/hypothenuse tan opposite side/adjacent side)


    Now you're given an angle - 42 degrees and one side that is opposite from the angle which is 12.8 but we want to find the length of the hypothenuse

    so the two sides involved in the equation are the opposite and the hypothenuse, this means we use sin.


    so set it out: Sin(42) = 12.8 / the length of BD (in this case x)

    multiply both sides by x to get xsin(42) = 12.8 - the right hand side cancels out side you're also dividing by x


    Now to get x by itself divide both sides by sin(42) - since sin(42) is already being multiplied they will cancel out, leaving 12.8/sin(42)


    And that's it. x = 12.8/sin(42) = 19. something
    And this.

    If you just explain them so I can do them tonight or tomorrow morning xx



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    (Original post by z_o_e)
    What about this question?


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    try use pythagorus.
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    (Original post by Lucky10)
    Best advice i could give DO PAST PAPERS its the best way to get better a maths. Any topics you dont get use sites like mathswatch.
    i agree with that
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    (Original post by z_o_e)
    And this.

    If you just explain them so I can do them tonight or tomorrow morning xx



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    (Original post by z_o_e)
    And this.

    If you just explain them so I can do them tonight or tomorrow morning xx



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     x=4.3 \cos 50^{\circ} .
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    (Original post by B_9710)
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     x=4.3 \cos 50^{\circ} .
    Might as well write the answer out while you're at it...
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    (Original post by 34908seikj)
    Might as well write the answer out while you're at it...
    Fine, the answer is
     \displaystyle \frac{116+43 \left (\sqrt[3]{-\frac{\sqrt 3 }{2}+\frac{1}{2} i } + \sqrt[3]{-\frac{\sqrt 3 }{2}-\frac{1}{2} i } \right )}{20} .
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    (Original post by 34908seikj)
    try use pythagorus.
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    A bit stuck.


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    (Original post by z_o_e)
    A bit stuck.


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    Use pythagorus

     \sqrt{x^2 + (2x)^2} = 25


     \sqrt{x^2 + 4x^2} = 25
 
 
 
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