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.

math  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

math  geekgirl95, Saturday, July 16, 2011 at 6:53pm
thanks :)

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

math  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 sineinverse, you get two answers, even though your calculator only gives you the angle in the first quadrant.
so the second angle is 18072.25 = 107.75

math  geekgirl95, Saturday, July 16, 2011 at 7:00pm
pls. explain

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