Question

I need help please medel for the person who can...

Answer 1

anyone?

Answer 2

http://www.purplemath.com/modules/sectors.htm
If you couldn't got it let me know to show how to apply on your example :)

Answer 3

@Omar_Elboredy can't you just explain it to me cause all i see is equations that i don't understand

Answer 4

find the are of the whole circle

Answer 5

find the are of the segment

Answer 6

area of the whole circle - area of segment

Answer 7

https://www.youtube.com/watch?v=ahOJtHBdBgE

Answer 8

so 60-8 ?

Answer 9

*60-18

Answer 10

https://www.mathsisfun.com/geometry/circle-sector-segment.html

Answer 11

i honestly don't understand why i just can't have it explained instead of being lead to all these links.

Answer 12

OK, w8 I will demonstrate to U step by step , give me 10 mins :)

Answer 13

thanks so much @Omar_Elboredy

Answer 14

First of all suppose the circle to a whole pizza and you took from it a 60 deg slice, calculate the area of the current part of the pizza which will be
\[A1 = \frac{ 360-60 }{ 2 } * \frac{ \pi }{ 180 } * 18^{2}\]
which will equal to 270π then add to it the area of triangle resulting from removing the curvy part off of the 60 deg slice which will be
\[A2 = \frac{ 18*18*\sin(60) }{ 2 }\] which is \[81\sqrt{3}\]
So our result will be by adding both A1 and A2 we just calculated as above ..
Then tht total area of the shaded part will be : \[A = A1 + A2 = (270\pi + 81\sqrt{3})m ^{2}\]

Answer 15

oh the problem makes a little since i really didn't know this was a sin cos tan problem, thanks so much (medels)