math

Triangle ABC has are 30cm^2, AC=9 cm and BC= 7cm. Calculate the measureof angle ACB and the larget possible length of AB.

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  1. Area = (1/2)(AC)(BC)sinØ , where Ø is angle ACB
    30 =(1/2)(9)(7)sinØ
    sinØ = 60/63 = .95238
    Ø = 72.25° or Ø = 107.75°

    The largest value of AB occurs when Ø = 107.75

    by cosine law
    AB^2 = 7^2 + 9^2 - 2(7)(9)cos 107.75
    = 168.4187
    AB = √168.4187 = 12.98

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    posted by Reiny
  2. thanks :)

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  3. how did yu get 107.75?

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  4. the second angle is in quadrant II
    the sine is positive in first ad second quadrants.
    so when you take sine-inverse, you get two answers, even though your calculator only gives you the angle in the first quadrant.

    so the second angle is 180-72.25 = 107.75

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    posted by Reiny
  5. pls. explain

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  6. oh yeah thaNKS.:)

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