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Math Help - Trigonometry Question... (Possibly Using the Cosine Rule)

  1. #1
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    Red face Trigonometry Question... (Possibly Using the Cosine Rule)

    Hello,

    In this question, I am having trouble imagining what the diagram should look like. Am I finding the height of the triangle? I am feeling a little vexed...

    An advertising balloon is attached to two ropes 120 m and 100 m long. The ropes are anchored to level ground 35 m apart. How high can the balloon fly?


    Many thanks!
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  2. #2
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    Quote Originally Posted by JadeKiara View Post
    Hello,

    In this question, I am having trouble imagining what the diagram should look like. Am I finding the height of the triangle? I am feeling a little vexed...

    An advertising balloon is attached to two ropes 120 m and 100 m long. The ropes are anchored to level ground 35 m apart. How high can the balloon fly?


    Many thanks!
    If the foot of the perpendicular from the balloon is x m from left end, then
    x^2 + h^2 = 120^2.........(1)
    (35 - x)^2 + h^2 = 100^2 .....(2)
    solve these two equations and find x. Substitute this value in eq. 1 to find h.
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  3. #3
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    Hmmm...

    As I am new to simultaneous equations, please correct me if I am wrong:

    x^2 + h^2 = 120^2 ....... (1)
    (35 - x)^2 + h^2 = 100^2 ....... (2)

    x^2 + h^2 = 120^2 ....... (1)
    35^2 -70x + x^2 + h^2 = 100^2 ....... (2)


    How do I proceed? Adding or subtracting x^2 or h^2 ?
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  4. #4
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    Quote Originally Posted by JadeKiara View Post
    As I am new to simultaneous equations, please correct me if I am wrong:

    x^2 + h^2 = 120^2 ....... (1)
    (35 - x)^2 + h^2 = 100^2 ....... (2)

    x^2 + h^2 = 120^2 ....... (1)
    35^2 -70x + x^2 + h^2 = 100^2 ....... (2)


    How do I proceed? Adding or subtracting x^2 or h^2 ?
    From eq.1 you get
    h^2 = 120^2 - x^2
    Put it in the second equation and solve for x.
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  5. #5
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    So this is, indeed, a simultaneous equation?

    Can I expand (35 - x) ^2? That is, I think, where I am going wrong.
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    Quote Originally Posted by JadeKiara View Post
    So this is, indeed, a simultaneous equation?

    Can I expand (35 - x) ^2? That is, I think, where I am going wrong.
    In the eq.2 you have already expanded the above term.
    35^2 -70x + x^2 + h^2 = 100^2 ....... (2)
    Now substitute h^2 = 120^2 - x^2 in the above equation.
    35^2 -70x + x^2 - x^2 + 120^2 = 100^2 ....... (2)
    35^2 -70x + 120^2 = 100^2 ....... (2)
    Now solve for x.
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  7. #7
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    Quote Originally Posted by JadeKiara View Post
    Hello,

    In this question, I am having trouble imagining what the diagram should look like. Am I finding the height of the triangle? I am feeling a little vexed...

    An advertising balloon is attached to two ropes 120 m and 100 m long. The ropes are anchored to level ground 35 m apart. How high can the balloon fly?


    Many thanks!
    To find the maximum height, using the cosine rule....

    The height is maximum when the ropes are stretched to their limit,
    forming a triangle.

    Then, if we find the angle opposite the 100 m rope

    100^2=35^2+120^2-2(120)(35)cos\theta

    2(120)35cos\theta=35^2+120^2-100^2

    cos\theta=\frac{35^2+120^2-100^2}{(70)(120)}

    \theta=arccos\left(\frac{35^2+120^2-100^2}{(70)(120)}\right)

    Then

    sin\theta=\frac{h}{120}\ \Rightarrow\ h=120sin\theta
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