Altitude of Plane

[nycmathguy]

Section 4.8

If my set up picture is wrong, can you please draw the correct picture leading to the equation(s) needed to find A.

 

[MathLover1]

 

[nycmathguy]

View attachment 1176

Thanks. How did you do that? Anyway, I will work on this problem on my next day off. Taking a small break from precalculus for a few days.

 

[nycmathguy]

View attachment 1176

I really tried to get through this one. Can you set up the equation(s) for me? I will then do the work. I can't figure out how to find distance 1 and 2. How does 550 mph play into finding the approximate altitude of the plane?

 

[HallsofIvy]

Your picture has the observer on the ground moving forward (at 550 mph?!). The observer is stationary and it is the jet that is moving. Your picture should have the two lines, at 15 and 67 degrees, together at the bottom and apart at the top.

Since the plane is flying at 550 mi per hour, in one minute it will fly 550/60= 9 and 1/6 miles.
That will be the length of the line connecting those two lines at altitude h.

 

[nycmathguy]

Your picture has the observer on the ground moving forward (at 550 mph?!). The observer is stationary and it is the jet that is moving. Your picture should have the two lines, at 15 and 67 degrees, together at the bottom and apart at the top.

Since the plane is flying at 550 mi per hour, in one minute it will fly 550/60= 9 and 1/6 miles.
That will be the length of the line connecting those two lines at altitude h.

MathLover1 corrected my horrible picture. However, even using her fabulous drawing, I just haven't been able to find distance 1 and 2. Can you set up the proper equation(s) needed to answer the question on my own?

 

[nycmathguy]

View attachment 1176

Can you do this one for me? I vainly-tried several times.

 

[MathLover1]

the plane is flying at 550 mi per hour, in one minute it will fly 550/60= 55/6 miles per 1 min

let altitude be y, then we have

tan(16 )= y/(x+55/6)
y=tan(16 )*(x+55/6)

tan(57 )= y/x
y=tan(57 )*x

since height is same

tan(16 )*(x+55/6)=tan(57 )*x
x=2.0976 -> distance after 1 min

substitute x in formulas above
y=tan(16 )*(2.0976+55/6)
y=3.23mil

y=tan(57 )*2.0976
y=3.23mil

 

[nycmathguy]

the plane is flying at 550 mi per hour, in one minute it will fly 550/60= 55/6 miles per 1 min

let altitude be y, then we have
View attachment 1197

tan(16 )= y/(x+55/6)
y=tan(16 )*(x+55/6)

tan(57 )= y/x
y=tan(57 )*x

since height is same

tan(16 )*(x+55/6)=tan(57 )*x
x=2.0976 -> distance after 1 min

substitute x in formulas above
y=tan(16 )*(2.0976+55/6)
y=3.23mil

y=tan(57 )*2.0976
y=3.23mil

You created two equations in two unknowns. Right? It is easy to see how topics learned back in algebra 1 and 2 pop-up often in later studies. I am going to try this problem again with new numbers.

My Altitude Problem:

600 mph
Angle of Elevation is 18° at one point and 76° 2 minutes later. Look for this set up, picture and math work later today in a separate thread.