radians c2

THE diagram shows a circle with centre O and radius 6cm. The chord PQ divides the circle into a minor segment R1 of are A1cm^2 and a major segment R2 of area A2cm62. the chord PQ subtends an angle theta radians at O.

b) prove that f(theta) = 0
c) evaluate f(2.3) and f(2.32) and deduce that 2.3<thera<2.32

thanks

What's f?

OP

dvs

What's f?

i know as mucha s you about f

another

The diagram shows the cross section ABCD of a glass prism. AD=BC=4cm and both are at right angles to DC. AB is the arc of circle, centre O and radius 6cm. given that angle AOB = 2 thera and that the perimeter of the cross section is 2(7 + pi) cm:

a) show that (2theta + 2 sin theta - 1) = pi/3

b) verify that theta = pi/6

c) find the area of the cross section

Reply 3

Dez

20

No, seriously, what is f? There's no mention of it in the question. What you've just asked is the equivelant of saying "There's five peaches in the basket. How many hours did I spend weaving the basket?"

Your second question lacks information too. What cross-section?

Reply 4

Michael Scofield

Or 'how many hours were spent growing the peaches?' could be another equivalent phrase.... :|

Reply 5

Dez

20

Quite. I expect some supplied diagram solves the mystery for him, but unfortunately there's no seeing it here.

Reply 6

Piggy

1

GrantMac

THE diagram shows a circle with centre O and radius 6cm. The chord PQ divides the circle into a minor segment R1 of are A1cm^2 and a major segment R2 of area A2cm62. the chord PQ subtends an angle theta radians at O.

b) prove that f(theta) = 0
c) evaluate f(2.3) and f(2.32) and deduce that 2.3<thera<2.32

thanks

Ah I just read this question at school - apparently you don't need to know how to do part C) cos its in C3, but for b) you find the area of the minor sector, take that away from pi r^2 to give the area of major sector and then put that equal to 3 x area of small sector (which you are given), get all terms to one side and you'll find that all that *junk* = 0. Hope that helps and I apologise if I wrote something wrong - I don't have my book with me :wink:

Reply 7

insparato

15

Grant hadnt given you the information you needed to do this here it is.

Given that A2 = 3A1 and f(theta) = 2theta - 2 sin theta - pi

prove that f(theta) = 0
evaluate f(2.3) and f(2.32) and deduce that 2.3<theta<2.32

3[18(theta- sin theta)] = 6^2pi - 1/2(6^2)(theta - sin theta)
54(theta - sin theta) = 36pi - 18(theta - sin theta)
72(theta - sin theta) = 36 pi
pi = 2(theta - sin theta)
pi = 2theta - 2sin theta
2theta - 2sin theta - pi = 0

If f(theta) = 2theta - 2 sin theta - pi --- therefore f(theta) = 0

c) sub theta = 2.3 into 2theta - 2sin theta - pi = 0

so 2(2.3) - 2 sin (2.3) - pi = -0.033 (3dp)
2(2.32) - 2sin (2.32) - pi = 0.034 (3dp)

thus 2.3<theta<2.32 because -0.033<0<0.034

f(theta) is just like a function of theta like f(x).

Reply 8

jpepsred

9

I don't think your answer to part c) is correct, insparato as the 2(sin(theta)) term could be negative or positive, depending on theta. Therefore it doesn't necessarily follow that if f(2.3) is negative then f(2.3) is smaller than f(theta).

EDIT: just realised how old this thread is. Guess I'll have to look elsewhere for help here.

(edited 9 years ago)

Reply 9

coconut64

16

Original post by jpepsred

I don't think your answer to part c) is correct, insparato as the 2(sin(theta)) term could be negative or positive, depending on theta. Therefore it doesn't necessarily follow that if f(2.3) is negative than f(2.3) is smaller than f(theta).

EDIT: just realised how old this thread is. Guess I'll have to look elsewhere for help here.

Hi how do you know that you need to minus the area of the minor segment from 36 pi? Because a2=3a1 so it is just 54(theta-sine theta), it makes sense but how do you know if you need to do that extra step? Thanks

radians c2

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