Q5 c)i)1) (2 Viewers)

[sinophile2]

How the hell do you find the length of chord BA???A?@??!@?@#?

 

[copeys]

How the hell do you find the length of chord BA???A?@??!@?@#?

I used cosine rule, hoping it's right.

 

[domo.kun]

I used the cosine rule

AB^2 = 2^2 + 2^2 - 2*2*2*cos(pi/3)
AB^2 = 4
AB = 2

 

[nada92]

I used the cosine rule

AB^2 = 2^2 + 2^2 - 2*2*2*cos(pi/3)
AB^2 = 4
AB = 2

WOOTT did that in the last like 15 seconds.

 

[Joustah]

The angles at the circumference had to be equal, since the triangle was isosceles (equal radius). But since the angle at the center was 60, then that meant the angles at the circumference were both 60 as well, therefore the triangle was equilateral. Therefore, the length was the same as the radius (2)

 

[Venom.]

Wasnt it just square root 3? Cuz of the 60; 30 triangles?

 

[Thecorey0]

The angles at the circumference had to be equal, since the triangle was isosceles (equal radius). But since the angle at the center was 60, then that meant the angles at the circumference were both 60 as well, therefore the triangle was equilateral. Therefore, the length was the same as the radius (2)

I was thinking that during it, seems bit of a paradox to me. Ended up with 2pi/3 +2 as the final answer for the perimeter though.

 

[AnandDNA]

lol its obviously equilateral

theta=pi/3 which is like 60 degrees and oa=ob thus it has to be equialteral

 

[ayehann]

I was thinking that during it, seems bit of a paradox to me. Ended up with 2pi/3 +2 as the final answer for the perimeter though.

i got that too but i simplified it to 2 ( pie/3 + 1)

 

[meilz92]

i also used the cosine rule

where b= 2
c= 2

and the angle for the cos bit was pi/3 or whateva it was they gave us in the quesstion

 

[TechnIx]

FUCK. I read the question wrong.
I just found the length of the arc AB, not the perimeter of the minor segment. GRR

 

[najib__nicholas]

cosine rule, x^2=4+4-2x2x2xcos60
x = 2
lol stupid time

 

[gorgioskit]

Wasnt it just square root 3? Cuz of the 60; 30 triangles?

No.

 

[oly1991]

yeh that stuff was given ay haha

but im not sure whether i got the part before it right....the part where it says to find the other value of theta, anyone one want to share what they got???

i think i might of got it, but ive got a little feeling that a may of missed something haha

 

[2009gibsor]

Wasnt it just square root 3? Cuz of the 60; 30 triangles?

I think i did that...

 

[boxhunter91]

Guys what was the value for theta? in the first part?

 

[x3.eddayyeeee]

other value for theta was 2pi/3

 

[absorber]

wasn't it length of arc + the 2 everyone keeps mentioning?

 

[fayad158]

will i get it rong if i put theta was equal to 7pie/3
when subbedin it still gives an area of root 3?

 

[Melisoverschool]

The angles at the circumference had to be equal, since the triangle was isosceles (equal radius). But since the angle at the center was 60, then that meant the angles at the circumference were both 60 as well, therefore the triangle was equilateral. Therefore, the length was the same as the radius (2)

this is what i did too