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 December 6th, 2015, 10:19 AM #1 Newbie   Joined: Nov 2015 From: brighton, uk Posts: 7 Thanks: 0 Calculating the radius & angle of sector. Hi All, wondering if anyone is able to help me solve this problem; Length of an arc of a circle is 6cm. The corresponding sector area is 150cm squared. Calculate i) the radius of the circle and ii) the angle subtended at the centre of the circle by the arc I found the radius by setting up a pair of simultaneous equations using arc length =r*theta and area of sector= 1/2*r^2*theta. By using these I calculated the radius to be 50cm which surely cannot be correct? Any help would be great thanks!! Last edited by wahhdoe; December 6th, 2015 at 10:37 AM.
 December 6th, 2015, 12:40 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,836 Thanks: 1479 $s = r \cdot \theta \implies \theta = \dfrac{s}{r}$ $A = \dfrac{1}{2}r^2 \theta$ $A = \dfrac{1}{2}r^2 \cdot \dfrac{s}{r}$ $A = \dfrac{1}{2}r \cdot s$ $r = \dfrac{2A}{s} = \dfrac{300}{6} = 50 \text{ cm}$ $\theta = \dfrac{6}{50} = \dfrac{3}{25} \text{ radians } \approx 7^\circ$
 December 6th, 2015, 06:52 PM #3 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Let me try : $\displaystyle \frac{6}{150}=\frac{2\cancel{\pi}r}{\cancel{\pi}r^ 2}$ $\displaystyle \frac{1}{25}=\frac{2r}{r^2}$ $\displaystyle \frac{1}{25}=\frac{2}{r}$ $\displaystyle r=2\times25$ r = 50 $\displaystyle \frac{\theta}{360^o}=\frac{6}{2\pi\times50}$ $\displaystyle \frac{\theta}{360^o}=\frac{6}{100\pi}$ $\displaystyle \frac{\theta}{2\cancel{\pi}}=\frac{3}{50 \cancel{\pi}}$ $\displaystyle \theta=\frac{3\times2}{50}$ $\displaystyle \theta=\frac{6}{50}$ $\displaystyle \theta=\frac{3}{25}$

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### sector of circle &angle

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