trigohelp!
posted by holly .
how do i do this question?
a ladder length 13m rests against a vertical wall with its foot on a horizontal floor at a distance of 5m from the wall. when the top of the ladder slips down a distance of x, the foot of the ladder moves out x. find the distance of x.
a^2 + b^2 = c^2
a^2 + 5^2 = 13^2 ==> 16925=144=a^2
therefore a = 12
after adjustment;
(12x)^2 +(5+x)^2 = 13^2
144  24x +x^2 + 25 + 10x +x^2 = 169
2x^214x=0
x(2x14)=0
x = 7 or 0
Therefore the distance moved is 7m.
thanks!
but shouldn't it be 0 or 7?
oh, sorry, i get it
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