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Find the altitude if the plane

Altitude The angles of elevation to an air plane from two points A and B on level ground are 71° and 88°,respectively .The points A and B are 5.5 miles apart and the airplaine is east if both points in the same vertical plane. Find the altitude of the plane. 

Comments

I feel that the problem has something in the data that is not totally correct, maybe I misunderstood it.
Francisco,
             If I understand the problem correctly, I feel that my solution is correct. Looking at your solution, you have 88º inside the triangle. Being an angle of elevation 88º should be outside the triangle if the plane is flying east. That makes the interior angle 92º, and therefore the third top angle should be 17º. Of course there could be something missing from the problem whereby we're all misunderstanding the problem.
Arthur D.
 
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3 Answers

If the plane and the two points are in the same vertical plane, the solution will be:
Calculate the angle formed by the visual lines at the plane point; It will be 180 - 88 - 71 = 21 degrees
According to the sinus law we may write:
5.5(dist between points)/sin 21 = a(opposite to angle A)/ sin 71 = b/sin 88
solving for a and b we obtain:
a = 14.51 miles
b = 15.33 miles
Then we can form tho rectangles triangles, for one the vertical side, forms 90 degrees with the soil then will be the altitude.
Sinus theorem; 14.1/1 = h/sin 88 solving for h = 14.49 miles
To check this answer we can do it in the other triangle:
15.33/1(sin 90) = h/ sin 71, solving for h gives 14.49. so CHECK!!!!!!!!!!!!
 

Comments

i belived this answer is correct because of the drawing i have of the problem
let point C be where the plane is
you have triangle ABC
angle A is 71º
angle B is 88º (outside triangle ABC)
angle B inside the triangle is 92º (180-88)
angle C is 17º (71+92+17=180)
using the law of sines:
sin(17)/5.5=sin(71)/BC
0.29237/5.5=0.94552/BC
0.29237(BC)=5.5(0.94552)
0.29237(BC)=5.20036
BC=5.20036/0.29237
BC=17.7869 miles
now you have another triangle CBD where CD is the altitude of the plane
using the law of sines again:
sin(90)/BC=sin(88)/CD where CD is the altitude
1/17.7869=sin(88)/CD
CD=17.7869(0.99939)
CD=17.77605 miles

Comments

USE THIS ONE!!  

(I didn't use law of Sines) ;-)

Excellent job Arthur!
ARTHUR, YOU ARE CORRECT IN YOUR ANSWER, MAYBE I MISUNDERSTOOD THE PROBLEM.
Thank you Francisco, and Pete as well, for checking my solution. I guess we got the correct solution. I think it is good that tutors look over other tutors' solutions and help each other. It all comes down to helping the students.
Arthur D.
am a little confuse with all the answers but actually in the drawing i have a triangle inside another triangle and 88 and 71 are inside not out side. 88 is inside of one of the triangles and 71 is also inside the second triangel
•  Plane
|
|
|
|
| altitude(a)
|
|
|
•________X______________•B______5.5m________•A
 
 
TanA=    a    
            x+5.5
 
TanB=   a  
           x
 
Tan(71°)(x+5.5) = a
Tan(88°)(x) = a
 
Tan(71°)(x+5.5)=Tan(88°)x
 
(x)Tan(71°)+(5.5)Tan(71°) = (x)Tan(88°)
 
(5.5)Tan(71°) = (x)Tan(88°)-(x)Tan(71°)
 
(5.5)Tan(71°) = x(Tan88°-Tan71°)
 
x =  (5.5)Tan71°   
     Tan88°- Tan71°
 
After finding x, plug into:   a = (x)Tan88°   or    a = Tan(71°)(x+5.5)     (the former is easier)
 
 
 

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