Figured out the answer? Find out if you have it right with the solution below...
Did you find the answer in a different way? What did you think of the problem? Let us know in the comments, and stay tuned for more Problems of the Week.
Monday, July 27, 2015
Problem of the Week
Here's a new Problem of the Week from the Center of Math, which is a bit harder than some other ladders-resting-on-walls problems you might know.
Figured out the answer? Find out if you have it right with the solution below...
Did you find the answer in a different way? What did you think of the problem? Let us know in the comments, and stay tuned for more Problems of the Week.
Figured out the answer? Find out if you have it right with the solution below...
Did you find the answer in a different way? What did you think of the problem? Let us know in the comments, and stay tuned for more Problems of the Week.
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I solved it by breaking up the problem into the 2 small triangles and calling the angle in the bottom right of both triangles A and the breaking up the ladder into 35-x and x. So sin(A)=12/x and cos(A)=12/(35-x). Then using sin^2(A) + cos^2(A) =1 getting the equation 176400-10080x-(937x^2)+(70x^3)-(x^4) which factorises into -(x-15)(x-20)(x^2-35x-588). I disregard the imaginary part and the -1 as i have no need for it in my answer. Then using Pythagoras on the smallest the triangle in the diagram, i get (y^2)+(12^2)=(x^2), y being the term that i want. I then sub in the factors, x=15 and x=20 and you get the values of y as 9 and 16 respectively.
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