SOLUTION: Dear Sir/Madam; A pleasant day! kindly help me with this word problem which involves right triangles. From the top of the mountain 40 meters above the sea land, the ang

Algebra ->  Trigonometry-basics -> SOLUTION: Dear Sir/Madam; A pleasant day! kindly help me with this word problem which involves right triangles. From the top of the mountain 40 meters above the sea land, the ang      Log On


   

Question 826184: Dear Sir/Madam;
A pleasant day! kindly help me with this word problem which involves right triangles.
From the top of the mountain 40 meters above the sea land, the angles of depression of two boats are 27° and 32° . What is the distance between the two ships if they are lying on a straight line with the observer on the mountain top?
Thank you very much for your help!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
This will be more understandable if you draw a diagram:
  • On the same horizontal line, label two points A and B. (These will be the two boats.)
  • Above and to the right of both A and B, label a point M. (This is the top of the mountain.)
  • Directly below M, on the horizontal line with A and B, label a point C.
  • Draw segment MC. Label its length as 40 meters (since this is the height of the mountain.)
  • Draw a horizontal line through M. (Angles of depression (and elevation for that matter) always have a horizontal for one side.)
  • Label a point Q to the left of M on this new horizontal line.
  • Angle QMA is the 27 degree angle of depression. But do not label it! Because the two horizontal lines are parallel, angle MAC is also 27 degrees. (They're alternate interior angles!) Label angle MAC as 27 degrees.
  • With the same logic, both angle QMB and MBC are 35 degrees. Label angle MBC.
  • Label the length of segment AB as "x".
  • Label segment BC as "y".
Note that the length of segment AC will be x+y.

We now have two right triangles. And in these two triangles we know two angles and the shared side, MC, of the two triangles. And we have expressions for the horizontal sides of these triangles, AC for triangle MAC and BC for triangle MBC. Since the vertical side, MC, is opposite to the known angles and since the horizontal sides are adjacent to the known angles, we can use tan (opposite/adjacent) for the equations:
tan%2827%29+=+40%2F%28x%2By%29 from triangle MAC, and
tan%2835%29+=+40%2Fy from triangle MBC.
With these two two-variable equations we can solve for x and y. I will be using the Substitution Method and I will solve the second equation for y:
tan%2835%29+=+40%2Fy
y%2Atan%2835%29+=+40
y+=+40%2Ftan%2835%29

Now I will substitute this in for the y in the first equation:
tan%2827%29+=+40%2F%28x%2By%29
tan%2827%29+=+40%2F%28x%2B%2840%2Ftan%2835%29%29%29
With only one variable left, x, we can solve for it. Multiplying both sides by the right side's denominator:
%28x%2B%2840%2Ftan%2835%29%29%29%2Atan%2827%29+=+40
Using the Distributive Property:
x%2Atan%2827%29%2B%2840%2Ftan%2835%29%29%2Atan%2827%29+=+40
Multiplying both sides by tan(35) [to eliminate the remaining fraction]:
x%2Atan%2827%29%2Atan%2835%29%2B40tan%2827%29+=+40tan%2835%29
Subtracting 40tan(27) from each side:
x%2Atan%2827%29%2Atan%2835%29+=+40tan%2835%29-40tan%2827%29
Dividing both sides by tan(27)*tan(35):
x+=+%2840tan%2835%29-40tan%2827%29%29%2F%28tan%2827%29%2Atan%2835%29%29
Since x is the distance between A and B (our boats), this is an exact expression for the answer to the problem.

But we probably want/need a decimal approximation. So we reach for our calculators to find the tan's:
x+=+%2840%280.7002%29-40%280.5095%29%29%2F%28%280.5095%29%2A%280.7002%29%29 [tan's rounded to 4 places]
Now we simplify:
x+=+%2828.008-20.38%29%2F%280.3568%29
x+=+%287.628%29%2F%280.3568%29
x+=+21.3789
So the boats are approximately 21 meters apart.