Posted by **geekgirl95** on Saturday, July 16, 2011 at 6:07pm.

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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**Reiny**, Saturday, July 16, 2011 at 6:36pm
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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**geekgirl95**, Saturday, July 16, 2011 at 6:53pm
thanks :)

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**geekgirl95**, Saturday, July 16, 2011 at 6:56pm
how did yu get 107.75?

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**Reiny**, Saturday, July 16, 2011 at 6:58pm
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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**geekgirl95**, Saturday, July 16, 2011 at 7:00pm
pls. explain

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**geekgirl95**, Saturday, July 16, 2011 at 7:01pm
oh yeah thaNKS.:)

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