SOLUTION: Apply the law of sines or the law of cosines and diagram and label the appropriate parts of the triangles. An airplane sights an airport runway at an angle of depression at 47

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Question 1150839: Apply the law of sines or the law of cosines and diagram and label the appropriate parts of the triangles.
An airplane sights an airport runway at an angle of depression at 47 degrees. After the plane flies 4 miles closer the angle of depression from the plane to the runway is 62 degrees. What is the air distance from the plane to the end of the runway at the second sighting?
Found 4 solutions by MathLover1, ikleyn, MathTherapy, jim_thompson5910:
Answer by MathLover1(20850) About Me  (Show Source):
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from triangle AOB we have:
tan%2847%29=h%2F4
=> h=4tan%2847%29.....eq.1

from triangle CDB we have:
tan%2862%29=h%2Fx
=> h=x%2Atan%2862%29.....eq.2

from eq.1 and eq.2 we have
4tan%2847%29=x%2Atan%2862%29
x=4tan%2847%29%2Ftan%2862%29
x=4%281.07236871%29%2F1.880726465346332
x=2.28mil

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.

As you see from the solution of the other tutor, NEITHER law of sines NOR law of cosines is relevant to the problem' solution.

Simple Trigonometry works and provides the solution.



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Apply the law of sines or the law of cosines and diagram and label the appropriate parts of the triangles.
An airplane sights an airport runway at an angle of depression at 47 degrees. After the plane flies 4 miles closer the angle of depression from the plane to the runway is 62 degrees. What is the air distance from the plane to the end of the runway at the second sighting? 
My understanding is that the problem is asking for the distance from the plane's height with a 62o angle of depression to runway, which, when using

MATHLOVER1's diagram, is side BD, or the hypotenuse of triangle BDC. And, side BD can be found by using the law of sines (as requested), as follows:  

matrix%281%2C3%2C+BD+%2A+sin+%2815%5Eo%29%2C+%22=%22%2C+4+%2A+sin+%2847%5Eo%29%29 ------- Cross-multiplying

Distance from the plane's height with a 62o angle of depression to runway, or 

FYI: When finding "h" in MATHLOVER1's diagram, and as she tried to, the ratio is NOT: matrix%281%2C3%2C+tan%2847%5Eo%29%2C+%22=%22%2C+h%2F4%29, but should either be: !!

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 11.3 miles
This answer is approximate and rounded to 1 decimal place.

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Explanation:

Diagram

The plane starts at point A, then it moves 4 miles to get to point B. Point C is the end of the runway location. Angles of depression are formed by starting off looking directly horizontally, then looking down at the target.

Segments
c = AB = 4
a = BC = x
lower case letters used for side lengths; upper case used for angles (note how lowercase 'a' is opposite uppercase 'A', same for B and C as well)
We don't need to find the length of segment b though it would be good practice for you to do later.

Angles
A = 47 degrees
B = 118 degrees (118 + 62 = 180)
C = 15 degrees (solve 47+118+y = 180 for y)

Law of Sines
sin(A)/a = sin(C)/c
sin(47)/x = sin(15)/4
sin(47)*4 = x*sin(15)
4*sin(47) = x*sin(15)
x*sin(15) = 4*sin(47)
x = 4*sin(47)/sin(15)
x = 11.302934856737
x = 11.3

The distance from point B to point C is roughly 11.3 miles