In the diagram in the left below, AD is a diameter of the circle and is tangent to line m at D. If AD=1 AND BC is perpendicular to m, show that:
a) AB + BC = 1 +cos(x) - cos^2(x)
b) What value of x makes the sum AB + BC a maximum?
In the diagram in the left below, AD is a diameter of the circle and is tangent to line m at D. If AD=1 AND BC is perpendicular to m, show that:
a) AB + BC = 1 +cos(x) - cos^2(x)
b) What value of x makes the sum AB + BC a maximum?
Hello, violetice!
$\displaystyle \Delta OAB$ is isosceles with $\displaystyle OA = OB = \tfrac{1}{2}$$\displaystyle AD$ is a diameter of circle $\displaystyle O$ and is tangent to line $\displaystyle m$ at $\displaystyle D.$Code:A * * * * |x * * * | * * * E + - - - -* B | * | * |θ * |* * O* |* * | |* | | * | * * | *| * | * | - - - - * * * - - -+- - - - m D C
If $\displaystyle AD=1$ and $\displaystyle BC \perp m$,
. . a) show that: .$\displaystyle AB + BC \:= \:1 +\cos x - \cos^2\!x$
Then: .$\displaystyle x \,=\, \angle OAB \,=\, \angle OBA \quad\Rightarrow\quad \angle AOB \,=\,\theta \,=\,180^o - 2x$
Law of Cosines: .$\displaystyle AB^2 \:=\:OA^2 + OB^2 - 2(OA)(OB)\cos\theta$
We have: .$\displaystyle AB^2 \:=\:\left(\tfrac{1}{2}\right)^2 + \left(\tfrac{1}{2}\right)^2 - 2\left(\tfrac{1}{2}\right)\left(\tfrac{1}{2}\right )\cos(180^o-2x) \:=\:\tfrac{1}{4} + \tfrac{1}{4} - \tfrac{1}{2}[-\cos2x] $
. . $\displaystyle AB^2 \:=\:\frac{1}{2}+\frac{1}{2}\cos2x \:=\;\frac{1+\cos2x}{2} \;=\;\cos^2\!x \quad\Rightarrow\quad\boxed{ AB \:=\:\cos x}$
We see that: .$\displaystyle BC \:=\:AD - AE$
In right triangle $\displaystyle AEB\!:\;\;AE \:=\:AB\cos x \:=\:\cos^2\!x$
Hence: .$\displaystyle \boxed{BC \:=\:1 - \cos^2\!x}$
Therefore: .$\displaystyle AB + BC \:=\:\cos x + 1 - \cos^2\!x$
b) What value of $\displaystyle x$ makes the sum $\displaystyle AB + BC$ a maximum?
We have: .$\displaystyle f(x) \:=\:1 + \cos x - \cos^2\!x$
Differentiate and equate to zero: .$\displaystyle -\sin x + 2\cos x\sin x \:=\:0$
. . Factor: .$\displaystyle \sin x(2\cos x - 1) \:=\:0$
We have: .$\displaystyle \sin x \:=\:0\quad\Rightarrow\quad x \:=\:0$ . This gives a minimum sum.
And: .$\displaystyle 2\cos x-1\:=\:0\quad\Rightarrow\quad \cos x \:=\:\tfrac{1}{2} \quad\Rightarrow\quad \boxed{x \:=\:\frac{\pi}{3}}$