Solve the problem that involves diameter of the Bullseye

[chwala]

Homework Statement

See attached.

Relevant Equations

angular method

I really do not understand what they are asking here...wording here in english is a bit confusing to me...but from similar examples, i made use of the approach below (which i still do not understand) hence my post.

$\dfrac{30π}{60 ×180} = \dfrac{0.0254}{d}$

$d = \dfrac{274.32}{30π}$

$d= 2.91$metres

 

[andrewkirk]

Given a line segment AB and a point P that lies on the perpendicular bisector of AB, we say that AB subtends an angle of $a$ at P if the angle $\angle APB=a$.
In this problem you have $\bar{AB}=0.0254$ (AB is any diameter segment of the bullseye), $a=30\ \mathrm{minutes}\ = 30\times \frac1{60}\times \frac{\pi}{180}\ \mathrm{radians}$ and you need to work out the distance $\bar{PX}$ where X is the midpoint of AB. You can do that exactly using trigonometry. Or you can use the approximation that, for small angles, which we have here, tan(a) approximately equals a. That is what your calculation does.
I get 2.91061 using the exact, trigonometric approach and 2.91063 using the small-angle approximation. So the approximation works well here.

 

[Lnewqban]

I really do not understand what they are asking here...wording here in english is a bit confusing to me...but from similar examples, i made use of the approach below (which i still do not understand) hence my post.

Please, see:

https://www.mathsisfun.com/definitions/subtended-angle.html