Asked by geekgirl95

Triangle ABC has are 30cm^2, AC=9 cm and BC= 7cm. Calculate the measureof angle ACB and the larget possible length of AB.
Answered by Reiny
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
Answered by geekgirl95
thanks :)
Answered by geekgirl95
how did yu get 107.75?
Answered by Reiny
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
Answered by geekgirl95
pls. explain
Answered by geekgirl95
oh yeah thaNKS.:)
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