Known AB=40km, AC=8√3km, ∠ BAC = 30+60 = 90, ∠ CAM = 90-60 = 30.
∴ BC=√(AB? +AC? )= 16√7
In the right triangle ABC:
cos∠ACB = AC/BC = 8√3/( 16√7)=√2 1/ 14
sin∠ACB = AB/BC = 40/( 16√7)= 5√7/ 14
Extend BC to Ann to D.
∠∠ACB =∠D+∠DAC
∴sin(∠d)= sin(∠ACB-∠DAC)= sin∠ACB * cos∠DAC-con∠ACB sin∠DAC
=5√7/ 14 * √3/2 - √2 1/ 14 * 1/2
=√2 1/7
sin∠ACD = sin( 180-∠ACB)= sin∠ACB
According to sine theorem
AD/sin∠ACD=AC/sin∠D
AD = 8√3/(√2 1/7)* 5√7/ 14 = 20k m
∫AM = 19.5,AN= 19.5+ 1=20.5
Am & ltAD & ltart 1 (before vowels and phonemes)
This ship can just reach the pier MN and the pier.