Question 42890
{{{drawing(600,600,-10,50,-10,50,
rectangle(0,0,2,40),
line(2,40,30,0),
line(-5,0,35,0),
locate(3,40,A),
locate(30.5,0,C),
locate(2.5,0,B),
locate(-7,20,ANTENNA),
locate(18,20,WIRE),
locate(20,-1,GROUND)
)}}}


Considering triangle ABC, <ABC is a right angle.
So we can apply Pythagorus theorem: {{{Hypotanuse^2 = Base^2 + Height^2}}}


Here base = BC, height = AB and hypotanuse = AC.


Given: AC = 50 ft.
Let BC = x feet. 
Then, AB = (x+10) feet 

[This is because height of the antenna is 10 ft larger than the distance from the base of the antenna to the point where the guy wire is attached to the ground].


So from Pythagorus theorem we have
{{{50^2 = x^2 + (x+10)^2}}}
or {{{50^2 = x^2 + x^2 + 20x + 10^2}}}
or {{{2x^2 + 20x + 10^2 - 50^2 = 0}}}
or {{{2x^2 + 20x + 100 - 2500 = 0}}}
or {{{2x^2 + 20x - 2400 = 0}}}
or {{{x^2 + 10x - 1200 = 0}}}
or {{{x^2 + 40x - 30x - 1200 = 0}}}
or {{{x(x + 40) - 30(x + 40) = 0}}}
or {{{(x+40)(x-30)=0}}}


So either (x + 40) = 0 or (x - 30) = 0 i.e. either x = -40 or x = 30.
Now, 'x' being the length of a side of a triangle cannot be negative.


So x = 30 i.e. the distance of the base of the antenna from the point where the guy wire is attached to the ground is 30 feet.


Hence, the height of the antenna is (x+10) = 30+10 = 40 feet.