A kite has 120m of string attached to it, when at an elevation of 60 degree. How far is it above the hand holding it?

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Answer 1 · 2018-03-22T05:28:56

The elevation is #103.92 or 60sqrt(3)# meters

The are two ways of approaching this. One is using the sine relationship, and the other is using a triangular identity.

The identity I'm talking about is the side ratios of a 30-60-90 triangle.

The short side (from the 60-degree to the 90) is ratio of 1.

The hypotenuse is ratio of 2

The long side (from the 90 to the 30-degree) is ratio of #sqrt(3)#

So, if the string length is 120, and is forming a 60-degree angle with the horizontal, the ratio of the hypotenuse to the altitude is:

#2:sqrt(3)#

Using this:

#"hyp"=2/sqrt(3)"alt" rArr 120=2/sqrt(3)"alt"#

#sqrt(3)/2xx120="alt"#

#color(red)("alt"=60sqrt(3)=103.92#

The other method is using sine. Since the side relationship of sine is:

#Sin(x)="opposite"/"hypotenuse"#

we can re-write this to get a direct relationship, knowing that the 'opposite' side is the altitude:

#Sin(60)="alt"/120 rArr 120xxSin(60)="alt"#

It turns out that #Sin(60)=sqrt(3)/2~=0.866#, which bolsters the triangle identity above... finishing it up:

#120sqrt(3)/2="alt"#

#color(blue)("alt"=60sqrt(3)=103.92)#