Question 1189731: While standing on a 75m tall bridge, given can see two boats. From his position on the bridge, the first boat is located on a bearing of 70 degrees and the second boat is located on a bearing of 300 degrees. Gavin estimates that the angles of depression for each of the boats are 38 degrees and 47 degrees respectively. How far apart are the boats?
Thanks
Found 3 solutions by ankor@dixie-net.com, math_tutor2020, greenestamps:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! While standing on a 75m tall bridge, given can see two boats.
From his position on the bridge, the first boat is located on a bearing of 70 degrees and the second boat is located on a bearing of 300 degrees.
Gavin estimates that the angles of depression for each of the boats are 38 degrees and 47 degrees respectively.
How far apart are the boats?
:
find the actual distance from the observer to each boat.
a right angle is formed from the water, 75' below the bridge. to the boat.
the distance to boat will be the hypotenuse (h).
Find the interior angle from the given depression, 90-47 = 43 degrees
Cos(43) = 75/h
h = 102.55 m to one boat
find the distance to the other boat interior angle = 90-38 = 52 degrees
cos(52) = 75/h
h = 121.82 m to the other boat
:
these form the two sides of a triangle from the observer to each boat, the angle at the observer: 300=70 = 230 degrees (A), distance between the boats is a
Use the law of cosines to find a
You do this tedious math yourself, I got
a = 203.5 m between the boats
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answer: 150.79 meters approximately
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Explanation:
Define these points- A = location of boat 1
- B = location of boat 2
- C = point on the water's surface directly below Gavin's location
- D = Gavin's location on the bridge
For the sake of simplicity, we'll assume points A,B,C are all on the same flat plane.
Realistically, the water is bobbing the boats around, so such an assumption isn't entirely valid (but again we'll make things relatively simple).
Due to the 3D nature of this problem, we'll have to break it down into 2D pieces.
On a separate part of the paper, draw out right triangle ACD. This is a side profile view for the boat 1.
Here are all the relevant drawings needed.
For now, focus on triangle ACD only.
That triangle has these sides
CD = 75 meters = height
AC = unknown
Use the tangent ratio to get...
tan(angle) = opposite/adjacent
tan(A) = CD/AC
tan(38) = 75/AC
AC*tan(38) = 75
AC = 75/tan(38)
AC = 95.99562 approximately
Notice how angle A of triangle ACD is equal to the angle of depression for the first boat.
This is due to the alternate interior angles theorem.
Through similar calculations for triangle BCD, you should find that BC = 75/tan(47) = 69.93863 meters approximately.
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Now we'll consider the birds-eye-view to look directly down on triangle ABC.
We already found that
AC = 95.99562
BC = 69.93863
which are sides b and 'a' in that order
In other words,
b = AC = 95.99562
a = BC = 69.93863
The ultimate goal is to find the length of segment c = AB, which is the distance between the boats.
The bearings 70 and 300 have a gap of 300-70 = 230 degrees between them.
The remaining bit is 360-230 = 130 degrees
This measures angle ACB, aka angle C.
Use the law of cosines to find side c = AB
c^2 = a^2+b^2-2*a*b*cos(C)
c^2 = 69.93863^2+95.99562^2-2*69.93863*95.99562*cos(130)
c^2 = 22,737.6687
c = sqrt(22,737.6687)
c = 150.79 meters is the approximate distance between the two boats
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The answer from @math_tutor2020 is correct; it uses the law of cosines on a triangle with its sides on the surface of the water, which is correct.
The answer from the other tutor forms a triangle using the distance between the boats and the line-of-sight distances from the observer (75m above the water) to the two boats; but the given bearings are measured at the surface of the water -- not from the observer.
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