Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My question is- From an aeroplane vertically over a straight road,the angles of depression of two consecutive kilometer-stones on the same side are 45 degrees and 60 degrees.Find the height of the aeroplane from the road.

Any solution to solve this question would be greatly appreciated.

share|improve this question
    
It sounds like you have a triangle of which you know all the angles, and just one of the sides. Could you apply the sine rule maybe, to get another of the sides? From there, it's just a single trigonometric identity to get the height of the plane. Also, if this is homework, you should probably add the "homework" tag to the question, in case anyone feels inclined to provide a complete solution. –  user22805 Sep 15 '12 at 7:04

2 Answers 2

up vote 1 down vote accepted

Let $A$ be the position of the airplane, and let $B$ be the point on the road directly below the plane. Let $C$ be the location of the nearer kilometre marker, and let $D$ be the position of the further one. Make a suitable labelled diagram.

We are told that $\angle BAC$ is $30^\circ$ ($60^\circ$ below the horizontal) and that $\angle BAD$ is $45^\circ$.

We have $$\frac{BC}{h}=\tan(30^\circ)=\frac{1}{\sqrt{3}}.$$

Similarly, $$\frac{BD}{h}=\tan(45^\circ) =1.$$

Thus $BC=\frac{h}{\sqrt{3}}$ and $BD=h$. But $BD-BC=1$. This gives the equation $$h\left(1-\frac{1}{\sqrt{3}}\right)=1,$$ and now we can solve for $h$.

share|improve this answer
    
Thanks for the help :) –  mgh Sep 15 '12 at 9:10

I assume this is homework, so I'll give you a hint.

Make a drawing and identify two triangles:

  • the first is a right triangle, with a horizontal leg on the street (say its length is $l$);
  • the second is an obtuse triangle, adjacent to the first, with a side on the street (consecutive to $l$), which is the distance between the two kilometer stones.

Both have the position of the airplane as their upper vertex, and their height will be the height of the airplane on the street, call it $h$.

Note that both $l$ and $L \equiv l+1$ can be expressed as a function of $h$ and of one angle of the triangles you drew. This leads to a simple equation for $h$.

share|improve this answer
    
After following the hints I got an answer please will you tell me if it is right? –  mgh Sep 15 '12 at 7:31
    
my answer is- height=(3 + sq root of 3)/2 –  mgh Sep 15 '12 at 7:32
    
That's right: by the way, my solution coincides with André's one. :) –  Andrea Orta Sep 15 '12 at 7:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.