Just to elaborate a little more, let’s assume the radius of the semi-circle and the hypotenuse length to be r and h resply. Now it’s easy to see that cos(ξ) = [r + r/sin(ξ)] /h or h = [r + r/sin(ξ)]/cos(ξ). Now setting dh/dξ =0 gives us (sin(ξ))^3+2(sin(ξ))^2 -1=0 and the only admissible root of this eqn. is sin(ξ)=(V5 -1)/2 Ajit
Good one! But there is also another solution with a quadratic equation, something like this:
sin(ξ)^2 + sin(ξ) = 1, giving the same solution. Getting this quadratic demands a little bit more trigonometry. You should start with the bi-sect of the other angle (90-ξ) and go from there. It is just beautiful.
http://i46.tinypic.com/2di3e9t.gif
ReplyDeletemin(tan(x)+1/tan(x)+1/cos(x))
sin(x)=(sqrt(5)-1)/2
newzad, how can you found the side:
ReplyDeletetan + 1/cos ?
Instead you can say: (1 + 1/sin) / cos
and go from there.
Just to elaborate a little more, let’s assume the radius of the semi-circle and the hypotenuse length to be r and h resply. Now it’s easy to see that cos(ξ) = [r + r/sin(ξ)] /h or h = [r + r/sin(ξ)]/cos(ξ). Now setting dh/dξ =0 gives us (sin(ξ))^3+2(sin(ξ))^2 -1=0 and the only admissible root of this eqn. is sin(ξ)=(V5 -1)/2
ReplyDeleteAjit
Good one!
ReplyDeleteBut there is also another solution with a quadratic equation, something like this:
sin(ξ)^2 + sin(ξ) = 1, giving the same solution. Getting this quadratic demands a little bit more trigonometry. You should start with the bi-sect of the other angle (90-ξ) and go from there. It is just beautiful.