Question 4266
My hat is off to Earlsdon, who apparently was solving this problem at the same time as I was, and who submitted the solution before I did!! 


I can't draw the triangle, but the kite string is 50 feet, which is the hypotenuse of the right triangle.  


Let x = the horizontal distance 
 x + 10 = the height of the kite (that even rhymes!!)

The legs of the right triangle are x and x+10, with the hypotenuse = 50.


By Theorem of Pythagoras, {{{a^2 + b^2 = c^2}}}

{{{x^2 + (x+10)^2 = 50^2}}}
{{{x^2 + x^2 + 20x + 100 = 2500}}}


This is quadratic, so combine like terms and subtract 2500 from each side of the equation to set it equal to zero:


{{{2x^2 + 20x - 2400 = 0}}}


Divide both sides by 2:
{{{x^2 + 10x - 1200 = 0}}}


Factor the trinomial:
{{{(x+40)(x-30) = 0}}}


x = -40 or x = 30


You can't have a negative side of a triangle, so reject the -40.  The answer is x = 30 for the horizontal distance, and x+10 = 40 feet for the height.


R^2 at SCC