# The shadow of a tree standing on a level plane is found to be 50 meters longer when sun's elevation is 30 degree than when it is 60 degree. Find the height of the tree.

Please refer the attached image(not to scale).

Let the height of the tree be h meters.

In triangle ABD,

`tan(30)=h/(x+50)`

`1/sqrt(3)=h/(x+50)`    -------- (1)

In triangle ABC,

`tan(60)=h/x`

`sqrt(3)=h/x`

`rArrx=h/sqrt(3)`

Substitute the above in equation (1),

`1/sqrt(3)=h/(h/sqrt(3)+50)`

`sqrt(3)h=h/sqrt(3)+50`

`h(sqrt(3)-1/sqrt(3))=50`

`h((3-1)/sqrt(3))=50`

`h(2/sqrt(3))=50`

`h=(50sqrt(3))/2`

`h=25sqrt(3)` m

`h~~43.30 m`

So, the height of the tree is `~~43.30 m`

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Let x be the length of the shadow when the sun's angle of elevation is 60 degrees. Then the length of the shadow when the sun's angle of elevation is 30 degrees will be x+50.

We assume that the tree is vertical so the base of the tree forms a right angle with the presumable level ground. The model then is a right triangle with one leg being h, the height of the tree. When the acute angle along the ground is 30 degrees the length of the remaining side is x+50; when 60 degrees it is 50.

Now in a 30-60-90 right triangle, the ratio of the legs of the triangle are sqrt(3) or its reciprocal.

When the sun's angle of elevation is 60 degrees we have the length of the shadow as h/sqrt(3) so h=sqrt(3)*x.

When the sun's angle of elevation is 30 degrees the length of the shadow is h*sqrt(3) so 2x+50=h*sqrt(3) or h=(x+50)/sqrt(30)

Substituting we get sqrt(3)x=(x+50)/sqrt(3)

3x=x+50

x=25

So the shadow's length is 25 feet when the sun is at 60 degrees, and 75 feet when the sun is at 30 degrees.

So the height of the tree is 25*sqrt(3) which is approximately 43.3m.

* If you know the tangent ratio:

tan(30)=h/(x+50) ==> h=tan(30)(x+50)
tan(60)=h/x ==> h=tan(60)*x

So tan(30)(x+50)=x*tan(60)

And (x+50)/sqrt(3)=sqrt(3)*x as before.

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