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The discus throw is an event in which an athlete attempts to throw a discus. The athlete spins anti-clockwise around one and a half times through a circle, then releases the throw. When released, the discus travels along tangent to the circular spin orbit.
In the given figure, $AB$ is one such tangent to a circle of radius 75 cm. Point $O$ is centre of the circle and $\angle ABO = 30^\circ$. $PQ$ is parallel to $OA$. Based on above information:
(a) find the length of $AB$.
(b) find the length of $OB$.
(c) find the length of $AP$.
OR
Find the length of $PQ$
In the given figure, $AB$ is one such tangent to a circle of radius 75 cm. Point $O$ is centre of the circle and $\angle ABO = 30^\circ$. $PQ$ is parallel to $OA$. Based on above information:
(a) find the length of $AB$.
(b) find the length of $OB$.
(c) find the length of $AP$.
OR
Find the length of $PQ$
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(i) $\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{75}{AB}$
$\Rightarrow AB = 75\sqrt{3}$ cm
(ii) $\sin 30^\circ = \frac{1}{2} = \frac{75}{OB}$
$\Rightarrow OB = 150$ cm
(iii) $QB = 150 – 75 = 75$ cm
$\Rightarrow Q$ is mid point. of $OB$
Since $PQ \parallel AO$ therefore $P$ is mid point of $AB$
Hence $AP = \frac{75\sqrt{3}}{2}$ cm.
OR
(iii) $QB = 150 – 75 = 75$ cm
Now, $\triangle BQP \sim \triangle BOA$
$\Rightarrow \frac{QB}{OB} = \frac{PQ}{OA}$
$\Rightarrow \frac{1}{2} = \frac{PQ}{75}$
$\Rightarrow PQ = \frac{75}{2}$ cm
$\Rightarrow AB = 75\sqrt{3}$ cm
(ii) $\sin 30^\circ = \frac{1}{2} = \frac{75}{OB}$
$\Rightarrow OB = 150$ cm
(iii) $QB = 150 – 75 = 75$ cm
$\Rightarrow Q$ is mid point. of $OB$
Since $PQ \parallel AO$ therefore $P$ is mid point of $AB$
Hence $AP = \frac{75\sqrt{3}}{2}$ cm.
OR
(iii) $QB = 150 – 75 = 75$ cm
Now, $\triangle BQP \sim \triangle BOA$
$\Rightarrow \frac{QB}{OB} = \frac{PQ}{OA}$
$\Rightarrow \frac{1}{2} = \frac{PQ}{75}$
$\Rightarrow PQ = \frac{75}{2}$ cm