posted 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.
Area = (1/2)(AC)(BC)sinØ , where Ø is angle ACB
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
AB = √168.4187 = 12.98
how did yu get 107.75?
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
oh yeah thaNKS.:)