Q) 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.

The discus throw_Circle CBSE 2023

In the given figure, AB is one such tangent to a circle of radius 75 cm. Point O is centre of the circle and ZABO =30″. PQ is parallel to OA.

The discus throw Circle important question CBSE 2023 NCERT

Based on above information:

(a) find the length of AB.

(b) find the length of OB.

(c) find the length of AP.


Find the length of PQ


(i) Value of line AB:

In ΔOAB, tan 30° = \frac{75}{AB}

\therefore         \frac{1}{\sqrt3}  = \frac{75}{AB}

AB = 75√3 cm

(ii) Value of line OB:

In ΔOAB, Sin 30° = \frac{75}{OB}

\frac{1}{2} = \frac{75}{OB}

OB = 75 x 2

OB = 150 cm

(iii) Value of line AP:

Now, given that radius OA = OQ = 75 cm

therefore, QB = OB – OQ = 150 -75 = 75 cm

Therefore, Q is midpoint of line OB

Given that PQ ǁ AO, and since, we just found that Q is midpoint of line OB,

Therefore, P is midpoint of AB.

Hence AP = \frac{AB}{2} = \frac{75\sqrt3}{2}

AP = \frac{75\sqrt3}{2} cm


Value of line PQ:

Method 1:

We have just found out that  OQ = QB = 75 cm

Now, In ΔQPB, Sin 30° = \frac{PQ}{QB}

\therefore      \frac{1}{2} = \frac{PQ}{QB}

PQ = \frac{QB}{2} = \frac{75}{2}

PQ = \frac{75}{2} cm

Method 2:


OQ = QB, AP = PB, ∠OBA = ∠QBP

Therefore, ΔOAB  ~ ΔQPB

\frac{QB}{OB} = \frac{PQ}{OA}

\frac{1}{2} = \frac{PQ}{75}

PQ = \frac{75}{2} cm

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