Question

There is a mark on the outermost part of a wheel of radius 30 centimetres. Now the mark is close to the ground as shown in the figure. If the wheel rolls 31.4 centimetres on a straight line, then
a) Find the angle by which the wheel rotates (use $\pi =3.14$ as an approximation).
b) What will be the height of the mark from the ground?

Answer 1

(a)The wheel rolls $31.4cm$ and radius of the circle is $30cm$
$s=31.4cm$ and $r=30cm$
The angle by which the wheel rotates
$=\frac{s}{2\pi r}\times {360}^o$
$=\frac{31.4}{2\times 3.14\times 30}\times {360}^o$
$={60}^o$
(b) In$△OAC$,
$\angle AOC={60}^o$
Also, $OA=OC$ [Radius of the circle]
$\therefore \angle OAC=\angle OCA$
$\angle OAC+\angle OCA+\angle AOC={180}^o$
$2\angle OAC={120}^o$
$\angle OAC={60}^o=\angle OCA$
$\therefore △OAC$ is an equilateral triangle.
$\therefore AC=30cm$
$\angle OAC+\angle CAB={90}^o$
$\angle CAB={30}^o$
In $△ABC$,
$\sin {30}^o=\frac{CB}{AC}$
$\frac{1}{2}=\frac{CB}{30}$
$\therefore CB=15cm$.