In curling, it is often necessary to hit and displace an opponent’s stone to win the end. Olivia would like to hit her opponent’s stone with her own stone. If she releases her stone at the hog line, it needs to travel another 87 feet before reaching her opponent’s stone. Olivia can be off by 11 inches in either direction and still hit her opponent’s stone. Assuming her stone doesn’t curl (change direction), within what angle, $\theta$, to the nearest tenth, must Olivia throw her stone to hit her opponent’s stone?
With this question, do you need to convert the 11 inches to feet, and then solve from there? And since both triangles are the same, do you only solve for one?
$11$ inches = $0.916667$ Feet
$\tan (a) = a / b$
$\tan (a) = 0.916667 / 87$
$a = \arctan (0.105364022989)$
$a = 0.6037^\circ$
Best Answer
There is something not quite right about this question. If we assume that
then you cannot actually come to the situation in the picture. The red stone would have to pass throug the yellow one, which is not very realistic. Nevertheless, the picture still shows a right angled triangle, with certain side lengths, so we can say that in the picture $\theta =\arctan \left (\frac{0.916667}{87}\right )\approx0.6037$.
Now I would like to actually solve what the maximum value for $\theta$ can be, for the red stone to still hit the yellow one, under the same assumtions as listed above.
Now I will drop all units and introduce
$B:=$ point on $l$, such that $d(A,l)=d(A,B)$, implying $\angle YBA=90^\circ$. We may obviously assume that $0\leq\theta < 90^\circ$. Note that $\sin (\theta)$ is stricly increasing on this interval.
Now, as $A'$ moves along $l$, the two stones will collide, if and only if $\vert AB\vert\leq 2r$. This means that $$ \sin (\theta) =\frac{\vert AB\vert}d \leq\frac{2r}d\implies \theta \leq \arcsin\left( \frac{2r}d \right).$$ And of course, the complete answer will be $\left\vert\theta\right \vert\leq \arcsin\left(\frac {2r} d\right) $.