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# A sector has an angle of 107.9 and an arc length of 5.92m. Find its radius

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A sector has an angle of 107.9 and an arc length of 5.92m. Find its radius

Guest Jan 14, 2015

#3
+18829
+10

A sector has an angle of 107.9 and an arc length of 5.92m. Find its radius

arc length  b = 5.92 m

angle  $$\alpha \ensurement{^{\circ}}$$ = $$107.9\ensurement{^{\circ}}$$

$$\boxed{b =r* \breve \alpha} \qquad \breve \alpha =\alpha \ensurement{^{\circ}} * \frac{2\pi}{360\ensurement{^{\circ}}} = \alpha \ensurement{^{\circ}} *\frac{\pi}{180\ensurement{^{\circ}}} \\\\ \begin{array}{rcl} b &=& r* \breve \alpha \\ &=& r * \alpha \ensurement{^{\circ}} *\frac{\pi}{180\ensurement{^{\circ}}} \end{array}\\\\\\ \boxed{r = \left( \frac{180\ensurement{^{\circ}} }{\pi} \right)*\frac{b} {\alpha\ensurement{^{\circ}}} = 57.2957795131*\frac{b} {\alpha\ensurement{^{\circ}}} }\\\\ \small{\text{  r = \left( \frac{180\ensurement{^{\circ}} }{\pi} \right)*\frac{5.92\ m}{107.9\ensurement{^{\circ}}} = 57.2957795131 * \frac{5.92\ m}{107.9\ensurement{^{\circ}}} = 3.14356825503\ m  }}$$

heureka  Jan 14, 2015
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#1
+91451
+10

A sector has an angle of 107.9 and an arc length of 5.92m. Find its radius

There are 360 degrees in a circle so we hve 107.9/360  of the circle here.

$$\\P=\frac{ 107.9}{360} *2\pi *r\\\\ 5.92=\frac{ 107.9}{360} *2\pi *r\\\\ r=\frac{5.92*360}{(107.9*2*\pi)}\\\\$$

Melody  Jan 14, 2015
#2
+81022
+10

A sector has an angle of 107.9 and an arc length of 5.92m. Find its radius

Let us convert 107.9 degrees to radians = 107.9 x pi / 180 = about 1.883 rads

And using  S = rΘ   where S= the arc length and Θ is in radians, we have

5.92m = r (1.883)     divide both sides by (1.883)

CPhill  Jan 14, 2015
#3
+18829
+10

A sector has an angle of 107.9 and an arc length of 5.92m. Find its radius

arc length  b = 5.92 m

angle  $$\alpha \ensurement{^{\circ}}$$ = $$107.9\ensurement{^{\circ}}$$

$$\boxed{b =r* \breve \alpha} \qquad \breve \alpha =\alpha \ensurement{^{\circ}} * \frac{2\pi}{360\ensurement{^{\circ}}} = \alpha \ensurement{^{\circ}} *\frac{\pi}{180\ensurement{^{\circ}}} \\\\ \begin{array}{rcl} b &=& r* \breve \alpha \\ &=& r * \alpha \ensurement{^{\circ}} *\frac{\pi}{180\ensurement{^{\circ}}} \end{array}\\\\\\ \boxed{r = \left( \frac{180\ensurement{^{\circ}} }{\pi} \right)*\frac{b} {\alpha\ensurement{^{\circ}}} = 57.2957795131*\frac{b} {\alpha\ensurement{^{\circ}}} }\\\\ \small{\text{  r = \left( \frac{180\ensurement{^{\circ}} }{\pi} \right)*\frac{5.92\ m}{107.9\ensurement{^{\circ}}} = 57.2957795131 * \frac{5.92\ m}{107.9\ensurement{^{\circ}}} = 3.14356825503\ m  }}$$

heureka  Jan 14, 2015
#4
+81022
+5

My answer and Melody's are approximately the same -  depending on the level of rounding. She has taken a "ratio" approach, while I have have used a "trig" approach.......same dog, different fleas....

And heureka has provided a very nice LaTex answer....utilizing a combination of both things....!!!

CPhill  Jan 14, 2015

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