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An arc AB of a circle with radius 28 cm and center O subtends an angle AOB at the centre. If the length of arc AB is $$\frac{88}{3}$$ cm, find the length of chord AB.

I haven't solve any question of this kind so no idea.

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    $\begingroup$ HINT: Use the arc length formula (i.e. $r\theta=l$ where $\theta$ is measured in radians) to calculate the angle AOB. You should then be able to calculate the length of the chord AB. $\endgroup$ – Mufasa Jan 3 '16 at 12:56
  • $\begingroup$ @Musafa I got AB=28 cm but the answer is 5 cm as shown by my book. $\endgroup$ – Ger Wyn Jan 3 '16 at 13:03
  • $\begingroup$ Please show your calculations so that we can help spot where you may have made a mistake. $\endgroup$ – Mufasa Jan 3 '16 at 13:04
  • $\begingroup$ $$\ theta$$=l/r =(88/3)/28=1.047 radians and converting it into degree, its approximately equal to 60°… And then using cosine law gives AB =28 cm $\endgroup$ – Ger Wyn Jan 3 '16 at 13:08
  • $\begingroup$ Your answer is therefore correct (approx $28.01$cm) and the book is incorrect $\endgroup$ – Mufasa Jan 3 '16 at 13:11
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The radian measure $\theta$ of a central angle of a circle with radius $r$ that is subtended by an arc of length $s$ is defined to be $$\theta = \frac{s}{r}$$ arc_length

Since you were given $s$ and $r$, you can solve for $\theta$.

You wish to find the length of chord $\overline{AB}$.

chord_length

Draw altitude $\overline{CD}$ to side $\overline{AB}$ of $\triangle ABC$, as shown in the diagram. Since $|AC| = |BC| = r$, $\overline{AC} \cong \overline{BC}$. $\overline{CD} \cong \overline{CD}$ by the reflexive property of congruence. Hence, $\triangle ACD \cong \triangle BCD$ by the Hypotenuse-Leg Theorem. Since corresponding angles of congruent triangles are congruent, $\angle ACD \cong \angle BCD$, so $\overline{CD}$ bisects $\angle ACB$. Hence, $m\angle ACD = \frac{\theta}{2}$. Thus, $$|CD| = r\cos\left(\frac{\theta}{2}\right)$$ and $$|AD| = r\sin\left(\frac{\theta}{2}\right)$$ Since corresponding sides of congruent triangles are congruent, $\overline{AD} \cong \overline{BD}$.

Since $|AB| = |AD| + |BD|$, we obtain $$|AB| = 2r\sin\left(\frac{\theta}{2}\right)$$

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Hint use $S=r\theta$ where $S=arc length$ and $\theta$ is central angle in radians. After that drop a perpendicular from central abgle to chord AB then caculate AB by using trigonometric ratios.

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  • $\begingroup$ Can it not be donee without dropping the perpendicular @ Archis $\endgroup$ – Ger Wyn Jan 3 '16 at 13:13

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