angle of flight

Jul 2008
26
0
I took off from an airport at an angle of 18° (nearest degree). At a horizontal distance of 1000m (3sf) away from where I take off there is a range of hills 300m (nearest 10m) high. If I maintain an angle of 18° show weather I will clear the range.

Please show how you answer this.
 
May 2008
20
4
Just to clarify, is the question basically asking will a plane taking off at an angle of 18degrees above the horizon make it over mountains 300m high that are 1000m away?

If so.

Find the minimum angle (theta) for which the plane flys over.
So tan(theta) = opposite / adjacent = 300 / 1000
(theta) = arctan (300/1000) = 16.7 degrees.

Since the minimum angle to fly over the mountain is 16.7 degrees, and the plane flys at an angle of 18 degrees.
Therefore it will clear it.
 
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Soroban

MHF Hall of Honor
May 2006
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Lexington, MA (USA)
Hello, 0110kim!

I took off from an airport at an angle of 18° (nearest degree).
At a horizontal distance of 1000 m away from where I take off
there is a range of hills 300 m high.
If I maintain an angle of 18°, show whether I will clear the range.

There are a number of ways to answer this question.
Code:
                              * B
                          *   |
                      *       |
                  *           | 300
              *               |
          * θ                 |
      * - - - - - - - - - - - *
      A          1000         C

The 300-m hill is 1000 m away.
What angle \(\displaystyle \theta\) is needed to clear the hill?

We have: .\(\displaystyle \tan\theta \:=\:\frac{300}{1000} \:=\:0.3\)

Hence: .\(\displaystyle \theta \:=\:\arctan(0.3) \:\approx\:16.699^o\)

Therefore, any angle \(\displaystyle \theta > 17^o\) will clear the range.





Code:
                              * B
                          *   :
                      *       :
                  *           : h
              *               :
          * 18°               :
      * - - - - - - - - - - - *
      A          1000         C

The plane takes off at 18°
How high will it be 1000 m away?

We have: .\(\displaystyle \tan18^o \:=\:\frac{h}{1000} \quad\Rightarrow\quad h \:=\:1000\tan18^o \:\approx\:325\)

The plane will be 325 m high; it will clear the range.


 
Jul 2008
26
0
thank you

Thank you so much...
 
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