#1

This is

And no divide by zeros.

I'm doing applications of trigonometry, and just finishing off the chapter. However...

The arc, AB, of a sector of a cirlce, centre O, radius 2cm, is 4cm long. Find;

a. the area of the sector AOB

b. the area of the triangle AOB

c. the area of the minor segment cut off by the line AB

With help, I've gotten the A. correct, but B is being a bastard as I can't find the triangle sides, and I haven't looked at C yet. Any help?

**revision**, not homework, don't let me hear any 'UG doesn't do homework lolols'.And no divide by zeros.

I'm doing applications of trigonometry, and just finishing off the chapter. However...

The arc, AB, of a sector of a cirlce, centre O, radius 2cm, is 4cm long. Find;

a. the area of the sector AOB

b. the area of the triangle AOB

c. the area of the minor segment cut off by the line AB

With help, I've gotten the A. correct, but B is being a bastard as I can't find the triangle sides, and I haven't looked at C yet. Any help?

#2

I remember doing this, but I can't remember any specifics.

Textbooks?

Textbooks?

#3

craaaaaap im not looking forward to C2 on monday =[

il get back to you if i work it out

il get back to you if i work it out

#4

AS Core for Edexcel, Pearson Longman. Page 289, question 3.

#5

the triangle is half length times brendth right? itd help if there was a picture...

#6

UG/0 = We're not doing your homework?

#7

the triangle is half length times brendth right? itd help if there was a picture...

The formale is 1/2absinC (when using radons). However, we don't have a value for a and b.

#8

first of all work in radians.

for the angle AOB, l=(theta)r

4=2(theta)

2=(theta)

for part b, the area of the triangle is A=1/2 abSinC

(a and b are the radii)

part c is then part a) - part b)#

Hope thats right- thats what I could think of offhand

you got it on monday?

I've got 9 exams between monday and wedesday =/

for the angle AOB, l=(theta)r

4=2(theta)

2=(theta)

for part b, the area of the triangle is A=1/2 abSinC

(a and b are the radii)

part c is then part a) - part b)#

Hope thats right- thats what I could think of offhand

you got it on monday?

I've got 9 exams between monday and wedesday =/

#9

well, C is A - B (the questions, not the points)

#10

A and B are the same as the radius, how silly of me.

Yeah, C2 on Monday. Last minute boost to try to get my C. C1, C2, and two history modules all on the same day :/

#11

Sweet, I've just started my revision on it now.

That's a tough day. Serves you right for taking history

That's a tough day. Serves you right for taking history

#12

The arc, AB, of a sector of a cirlce, centre O, radius 2cm, is 4cm long. Find;

a. the area of the sector AOB

b. the area of the triangle AOB

c. the area of the minor segment cut off by the line AB

a. Area of AOB = (1/2)(r)^2(O)

O = Theta, ok? So just substitute r=radius, and theta you find by doing:

Arc length = r(O). So 4=2(O), solve the simple equation. O = 2, yeah?

So Area AOB = 1/2(2)^2(2)

b. You have two sides of the triangle AOB, because they're both radii, so they're both r, which equals 2. Area of a triangle = 1/2ab(sin C) where C is the angle contained by two sides, here O. Sub values a and b and C in, 2, 2, and sin O respectively. Remember to put your calc into radians mode for sine 2, otherwise it won't work (?)

If I'm wrong...sorry.

C. Is simply the area of the sector minus the area of the triangle.

This would be a lot easier if I could see the diagram, and i'm too lazy to draw one for myself here. Thanks for the mental workout though.

#13

Yeah, C2 on Monday. Last minute boost to try to get my C. C1, C2, and two history modules all on the same day :/

yeah same bro, but I'm retaking the first history module, instead of C1

then tuesday off, wednesday I've got 3 latin, then 2 economics- takes the fucking michael, to be honest.

only had m1 before half term.

#14

yeah same bro, but I'm retaking the first history module, instead of C1

then tuesday off, wednesday I've got 3 latin, then 2 economics- takes the fucking michael, to be honest.

only had m1 before half term.

Amen. I've been working solidly this week, I wish it was more spread out.

Hope the exams do you well!

#15

Amen. I've been working solidly this week, I wish it was more spread out.

Hope the exams do you well!

haha I second your sentiments.

Yeah, good luck mate.

#16

I don't like working with radians, I don't know..

First find the perimetre of the circle, and divide it by 4 cm (the arc)....

You get 3.14 (pi)

Now, find the area of the circle, and divide it by the fraction that gave the before division. The area would be:

2^2*3.14

You get 4 cm2

that makes you find the area of the sector I think....

Now, have 360/the fraction (3.14), to find the angle of the triangle (114,6). Afterwards, use the Sine Law (Or consine one, I can't remember) to find the side AB of the triangle, now find the apothem of it (divide AB in 2, and use Pythagoras Theorem) and find the area of the triangle.

Fod C I suppose you just have area of sector-area of triangle right?

May not be right, as I didn't use radians (angles instead), but still...

First find the perimetre of the circle, and divide it by 4 cm (the arc)....

You get 3.14 (pi)

Now, find the area of the circle, and divide it by the fraction that gave the before division. The area would be:

2^2*3.14

You get 4 cm2

that makes you find the area of the sector I think....

Now, have 360/the fraction (3.14), to find the angle of the triangle (114,6). Afterwards, use the Sine Law (Or consine one, I can't remember) to find the side AB of the triangle, now find the apothem of it (divide AB in 2, and use Pythagoras Theorem) and find the area of the triangle.

Fod C I suppose you just have area of sector-area of triangle right?

May not be right, as I didn't use radians (angles instead), but still...

#17

Do you have the answers to this because I've worked it out but I'm not sure if I'm right.

#18

Radians

a) (radius^2*theta radians)/2= area

arc length = radius*theta

b) (1/2)*2^2 sin theta (from (1/2)absinC )

c) a) - b)

Do it all in radians! Make sure you're calculator is set to radians, not degrees.

Just sub in the numbers into the equations above and you'll have your answers. draw a diagram so you know what you're trying to find. (that's what i just did xD)

a) (radius^2*theta radians)/2= area

arc length = radius*theta

b) (1/2)*2^2 sin theta (from (1/2)absinC )

c) a) - b)

Do it all in radians! Make sure you're calculator is set to radians, not degrees.

Just sub in the numbers into the equations above and you'll have your answers. draw a diagram so you know what you're trying to find. (that's what i just did xD)

*Last edited by OddOneOut at May 28, 2008,*

#19

I'll write down what I have worked out;

A) S = r$ (where $ = theta)

Therefore $ = S/r

$ = 2

Area = 0.5*r^2$

So A = 4cm^2

B)Using the Cosine Rule (a^2 = b^2 + c^2 - 2bcCosA)

a^2 = 8 - 8cos2

a = 3.366cm (that is the third side of the triangle)

However, I don't know why I worked that out because it isn't really needed.

A = 0.5abSinC

Area = 2*Sin2

= 1.8186cm^2

C) 4 - 1.8186

= 2.1814cm^2

A) S = r$ (where $ = theta)

Therefore $ = S/r

$ = 2

Area = 0.5*r^2$

So A = 4cm^2

B)Using the Cosine Rule (a^2 = b^2 + c^2 - 2bcCosA)

a^2 = 8 - 8cos2

a = 3.366cm (that is the third side of the triangle)

However, I don't know why I worked that out because it isn't really needed.

A = 0.5abSinC

Area = 2*Sin2

= 1.8186cm^2

C) 4 - 1.8186

= 2.1814cm^2

#20

^ That's what I got.

Well, exept I didn't actually work anything out after I found out the radians. But it's what I would've done :P

Well, exept I didn't actually work anything out after I found out the radians. But it's what I would've done :P

#21

^ That's what I got.

Well, exept I didn't actually work anything out after I found out the radians.But it's what I would've done :P

Haha, I'll take your word for it