How Far Apart Are the Two Boats?

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let right angle be point A

given:
AP=2500m
angles of elevations are 32 degrees and 44 degrees

since we have elevations, we know that angle of depression B2 is also 33 degrees (by def of alternate interior angles)

same rule apply for B1, so B1=44

let the distance AB1 =x and B1B2 =d

then

tan(44)= 2500m/x
x=tan(44)*2500m
x=2414.22m

from smaller triangle we have

tan(33)= 2500m/(x+d)
(x+d)=2500m/tan(33)
2414.22m+d=3849.66m
d=3849.66m -2414.22m
d=1435.44 m
the distance between two boats is 1435.44m
 
let right angle be point A

given:
AP=2500m
angles of elevations are 32 degrees and 44 degrees

since we have elevations, we know that angle of depression B2 is also 33 degrees (by def of alternate interior angles)

same rule apply for B1, so B1=44

let the distance AB1 =x and B1B2 =d

then

tan(44)= 2500m/x
x=tan(44)*2500m
x=2414.22m

from smaller triangle we have

tan(33)= 2500m/(x+d)
(x+d)=2500m/tan(33)
2414.22m+d=3849.66m
d=3849.66m -2414.22m
d=1435.44 m
the distance between two boats is 1435.44m

I should have been able to do this problem after completing precalculus with you. I wonder if I truly learned anything in that course.
 

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