math
posted by geekgirl95 on .
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
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 
thanks :)

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 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 
pls. explain

oh yeah thaNKS.:)