elementary functions mat 114

Question

A box kite is flying at the end of a fixed length of string. The angle of elevation of the kite string is 56◦. A gust of wind lifts the kite 6 feet higher and the angle of elevation increases to 61◦. How long is the kite string? 

Answer 1

Answer 2

side "a" is 6'. α=5º. β=46º=(180-134).

So a/(sinα)=b/(sinβ)=c/(sinγ).

6/sin5=c/sin129

68.84=c/.777

c=53.343..........'

Remember; the wind is blowing toward you.

Answer 3

This problem is NOT elementary!

Draw a figure...and think carefully about it.

Let the point of origin of the kite string be labelled O and the end of the string in the first position be A and the end at the second position be B. Notice that B is to the left and above A.

Draw a horizontal from A intersecting the 2nd kite string and call this point C.

Angle AOB =5° and is the vertex angle of an isosceles triangle with base angles 87.5°.

Angle CAO is 56° (alternate-interior angles of parallel lines) and that makes angle CAB=31.5°

The distance from B to the horizontal AC is 6'.

Therefore AB=6/sin 31.5°=11.483285.

AB is the base of an isosceles triangle with vertex angle 5°.

The kite string is the length of the side of this isosceles triangle which is 1/2 of AB divided by the sin 2.5°...which 131.630518'.

I plotted this problem on DESMOS and believe this is the correct answer.

If you find a different answer, I would like to see it.