Trig function help. (1 Viewer)

[RHINO7]

Yes these two questions are simple but i can't work them out-

i)The area of the sector of a circle that is subtended by an angle of pie/3 at the centre is 6(pie)m^2. Find the radius of the circle.

ii) A circle has a circumference of 185mm. Find the area of the sector cut off by an angle pie/5 subtended at the centre.

Please help. Also, what is the formulae for surface area and volume of a cone.

 

[Forbidden.]

lol pie ... anyway ...

1) Substitute your given values into A_sector = 0.5r²θ and make r the subject.

Answer:

Spoiler

A_sector = 0.5r²θ
6π = 0.5 x r² x θ
6π = πr²/6
36π = πr²
r² = 36

.: r = 6 m

2) I'm sure you know the formula for the circumference of the circle is, right ?

Spoiler

C = 2πr

Make r the subject from the circumference of the circle, and use that to find the area of the sector with your given values.

Spoiler

C = 2πr
185 = 2πr
r = 185/2π

A_sector = 0.5r²θ
A_sector = 0.5 x (185/2π)² x (π/5)
A_sector = (34225/40π)mm²

OR if not asked for exact answer.

A_sector = 272.3538964 .... (by calc.)
A_sector = 272 mm² (to whole number)

Surface Area of Cone: S = πr(r + l)
Volume of Cone: (πr²h)/3

 

[RHINO7]

Thanks. And yes I did and do know C = 2πr.