1. ## Trig. segmet area+perimiter

Hi again, i'm currently studying for my pure maths intermediate level exam and i need some help with a trigonometry question please:

-------------

The points A and B lie on the circumference of a circle center O and radius

r cm, such that the angle AOB = θ
radians. The sector AOB has area 9cm^2 and the total perimeter of sector AOB is 15cm.

a)Show that r satisfies the equation 2r^2 - 15r + 18 = o
b)Find the value of
θ,explaining why there is only one possible value.

------------------
I know that area of sector = (1/2).(r^2).(θ
)
not sure if perimiter of sector would be length of arc

Any help is appreciated,thanks!

*sry i meant sector not segment in thread title

2. Originally Posted by floyd
Hi again, i'm currently studying for my pure maths intermediate level exam and i need some help with a trigonometry question please:

-------------

The points A and B lie on the circumference of a circle center O and radius

r cm, such that the angle AOB = θ
radians. The sector AOB has area 9cm^2 and the total perimeter of sector AOB is 15cm.

a)Show that r satisfies the equation 2r^2 - 15r + 18 = o
b)Find the value of
θ,explaining why there is only one possible value.

------------------
I know that area of sector = (1/2).(r^2).(θ
)
not sure if perimiter of sector would be length of arc

Any help is appreciated,thanks!

Yes, all looks good.

Remember too that the length of the arc is

$\frac{\theta}{2\pi}\cdot 2\pi r = \theta r$.

So the perimeter will be $\theta r + 2r = r(\theta + 2)$.

3. Originally Posted by Prove It
Yes, all looks good.

Remember too that the length of the arc is

$\frac{\theta}{2\pi}\cdot 2\pi r = \theta r$.

So the perimeter will be $\theta r + 2r = r(\theta + 2)$.
ok thanks, any idea how i can relate those two equations to solve a) and b)?

I factorised 2r^2 - 15r + 18 = o which gives r=6,r=3/2 but i still have no idea how to continue

4. Originally Posted by floyd
ok thanks, any idea how i can relate those two equations to solve a) and b)?

I factorised 2r^2 - 15r + 18 = o which gives r=6,r=3/2 but i still have no idea how to continue
You know the area and perimeter, so solve the equations simultaneously for $r$ and $\theta$.

Then substitute $r$ into the equation and see if it holds true.

5. Originally Posted by Prove It
You know the area and perimeter, so solve the equations simultaneously for $r$ and $\theta$.

Then substitute $r$ into the equation and see if it holds true.
Ok got it! This is what i did just in case some1 has same problem:

the two equations are
1)15 = r(θ+2)
2)9 = 1/2r^2θ

from 1) i get θ = 15/r -2 and i substitute it in 2)
which gives 9= 1/2r^2(15/r - 2)
9= 1/2(15r - 2r^2)
18= 15r - 2r^2
2r^2 - 15r + 18 = o

part b) by factorising u get r = 6 r = 3/2

using 1) 15= 6(θ+2) 2) 15= 3/2(θ+2)
θ = 1/2 θ = 8

------
Thanks,Prove it
The last thing I want to ask regarding question b) is why is there only one possible value of θ?

6. Can you have an angle greater than $2\pi$ radians?

7. Originally Posted by Prove It
Can you have an angle greater than $2\pi$ radians?
ah i see, so that means i choose the 1/2. Ok thanks!