This week's CR4 Challenge Question:
A goat is placed in a fencedin circular field with radius r. It is tethered to a point on the perimeter of the field by a rope which allows it to graze as far as the radius R. What is the value of R to the nearest cm, if the goat can graze exactly ½ the circular field area, and r = 10m
Thanks to SlideRuler for this question!
And the Answer Is....
The grazing area is segment ABC + segment ABD
Therefore ½*r²*(2θsin(2θ))+½*R²*(2Φsin(2Φ))=½*π*r². ... eq(1)
also we have θ+Φ+Φ = π (isosceles Δ origin,C,A) ... eq(2)
and r*sin(θ) = R*sin(Φ) (both equal to ½*AB) ... eq(3)
Replacing 2Φ with (θπ); and replacing R with r*sin(Φ)/sin(θ); and using the relationship sin²(Φ) = ½*(1cos(2Φ)) = ½*(1cos(πθ)) = ½*(1+cos(θ)) in equation (1) gives
(πθsin(θ))+(2θsin(2θ)π)*(1+cos(θ)/(2*sin²(θ))=0 ... eq(4)
This has only one variable θ, and the root can be found using Newton's approximation. x_{n+1} = x_{n}  y_{n}/y'_{n}
Using a starting value of θ=1 gives successively
θ =1 > 1.2143 > 1.2327 > 1.2354 > 1.2358 > 1.2359
and the corresponding values of R by using eq(2) to get Φ and eq(3) to get R; (remembering r=10m) are
R= 10m > 11.41m >11.56m > 11.58m > 11.59m > 11.59m
The answer to the given problem is therefore 11.59m

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