(Triangle ABC is scalene) AC is a straight shore line and B is a boat out at sea. Angle A is 60 degrees and Angle C is 45 degrees. Find the shortest distance from the boat to the shore if A and C are 5km apart...I honestly have no idea where to start with this. I drew it out, but it's not a right triangle, so I can't use sine, cosine, or tangent...

Answer 1

If the point in question is D, then we have

AD+DC = 5

BD/AD = tan 60°
BD/DC = tan 45°

so, we have

BD/√3 + BD/1 = 5
Now just solve for BD, the desired distance.

Answer 2

First of all, my approach was to use sine rule. We all know that DB would be the shortest distance. In order to find that, we need to know either AB or BC,

Answer 3

First of all, my approach was to use law of sine. We all know that DB would be the shortest distance. In order to find that, we need to know either AB or BC,

so i find AB, AB = (sin45/ sin 75) x 5 = 3.66 ,
and then we find BD

sin 60 = BD/AB
BD= 3.66 x sin60 = 3.17km ( Approximately)