Sunday

April 20, 2014

April 20, 2014

Posted by **sophie** on Sunday, May 22, 2011 at 2:34pm.

- math -
**bobpursley**, Sunday, May 22, 2011 at 3:21pmlengths of a right triangle? You mean the three sides?

Try it: lengths a, ar, ar^2

now see if Pythagoras postulate will work:

c^2=a^2 + b^2

Either a, or ar^2 is c. Try a first

a^2=(ar)^2+(ar^2)^2

dividing by a^2

1=r^2 + r^4

r^4+r^2-1=0

r^2 = (1+-sqrt(5))/2

r= 1.272

so sides are a; 1.272a; 1.618a

which means a was not the longest side.

check:

(1.618^2)=1^2+1.272^2

acute angles?

sinA= 1.272/1.618 Then B: sinB=1/1.618

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