shooterman
Junior Member
- Joined
- Aug 20, 2009
- Messages
- 57
There is a women on top of a light house that is 2800 ft tall and she sees a yacht and a barge in the distance.
She measured the angle of depression of the yacht to be 20 degrees and the angle of depression of the barge to be
15 degrees and 30 min. She thinks the yacht and the barge to be 300 ft apart. If they are less than 300 ft apart than she would sound the horn. Are they closer than 300ft? and will she sound the horn.
So i drew the problem to make it visual.
[attachment=0sx6kpk0]th_Problem.jpg[/attachmentsx6kpk0]
To figure how far the yacht and barge are apart i solved for x which is the distance from light house for each triangle with this equation.
Then subtract the distance of the big triangle from small.
sin 20 Degrees = 2800/x for the small triangle and for the big triangle sin 15 degrees 30'= 2800/x
Am i puting it in the correct equation?
I got 58 ft as an answer.
She measured the angle of depression of the yacht to be 20 degrees and the angle of depression of the barge to be
15 degrees and 30 min. She thinks the yacht and the barge to be 300 ft apart. If they are less than 300 ft apart than she would sound the horn. Are they closer than 300ft? and will she sound the horn.
So i drew the problem to make it visual.
[attachment=0sx6kpk0]th_Problem.jpg[/attachmentsx6kpk0]
To figure how far the yacht and barge are apart i solved for x which is the distance from light house for each triangle with this equation.
Then subtract the distance of the big triangle from small.
sin 20 Degrees = 2800/x for the small triangle and for the big triangle sin 15 degrees 30'= 2800/x
Am i puting it in the correct equation?
I got 58 ft as an answer.