Question

A wheel with diameter AB touches the horizontal ground at the point A. There is a rod BC fixed at B such that ABC is vertical. A man from a point P on the ground, in the same plane as that of wheel and at a distanced from A is watching C and finds that its angle of elevation is $\alpha$. The wheel is then rotated about its fixed centre O such that C moves away from the man. The angle of elevation of C when it is just about to disappear is $\beta$. Find the radius of the wheel and the length of the rod. Also find distance PC when C is just to disappear.

Answer 1

Know quantities are d, $\alpha ,\beta$. Required quantities are radius $\alpha$, road BC = x and PC
$BC=rod=x$ $\therefore AC=2\alpha +x$
The wheel touches ground at A and P is the point of observation such that AP = d. The elevation of C from P is $\alpha$
$\therefore \tan \alpha =\frac{2\alpha +x}{d}$
$\therefore x=d\tan \alpha -2\alpha$ ...(1)
Now the wheel is rotated so that rod BC becomes B'C' and C' is just on the point of disappearing as seen from P so that PC' is a tangent to wheel at L. The angle of elevation C' from P is given to be $\beta$. Now PL = PA as tangent from P are equal and OA = OL = $\alpha$
$\therefore \Delta POL=\Delta POA$
and OP bisects and angle $\beta$.
$\therefore \tan \frac{\beta }{2}=\frac{a}{d}$
$\therefore a=d\tan \frac{\beta }{2}$ ...(2)
$\therefore x=d(\tan \alpha -2\tan \frac{b}{2})$ by (1) and (2) ...(3)
above gives the radius of wheel and length x of rod in terms of known quantities $d,\alpha ,\beta$.
$P{C}^{\prime }=PL+L{C}^{\prime }$
$=d+\sqrt{\left(\right(x+a{)}^2-a^2}=d+\sqrt{(x^2+2ax)}$
$=d+\sqrt{\left(x\right(x+2a\left)\right)}$
$=d+\sqrt{(d\tan \alpha .d(\tan \alpha -2\tan \frac{\beta }{2}))}$
$=d+d\sqrt{({\tan }^2\alpha -2\tan \alpha \tan \frac{\beta }{2})}$ by (2) and (3)