• Jan 10th 2007, 01:43 AM
Kim2425
Ok, I have some questions, these are all using radians

Question 1
a) 20 degrees
b) -72 degrees
c) 400 degrees
d) -140 degrees
e) 760 degrees

Question 2
An angle "p" subtends an arc of length 25cm in a circle of radius R cm.
The area of the sector POQ is 72cm (squared)
Forumulate two equations in r and "p" Find the values of r and "p"

Question 3
A cylindrical pipe of diameter 1.5m contains water to a depth of 0.9m.
a) find the cross sectional area of the water
b) if the water is flowing at a rate of 60 litres per second fin dthe speed of the water in m/s

I would really appreciate any help offered, these are the final pieces of some homework and i really need help, thank you
• Jan 10th 2007, 05:41 AM
topsquark
Quote:

Originally Posted by Kim2425
Question 1
a) 20 degrees
b) -72 degrees
c) 400 degrees
d) -140 degrees
e) 760 degrees

The relationship between degrees and radians is that $\pi$ rad = 180 degrees.

a) $\frac{20 \, degrees}{1} \cdot \frac{\pi \, rad}{180 \, degrees} \approx 0.349 \, rad$

The others are done the same way.

-Dan
• Jan 10th 2007, 06:10 AM
Soroban
Hello, Kim!

You're expected to know these formulas.

In a circle of radius $r$ with a central angle of $\theta$ radians:

. . the length of arc is: . $s \:=\:r\theta$

. . the area of the sector is: . $A \:=\:\frac{1}{2}r^2\theta$

Quote:

2) An angle $\theta$ subtends an arc of length 25cm in a circle of radius $R$ cm.
The area of the sector POQ is 72 cm².
(a) Formulate two equations in $R$ and $\theta.$
(b) Find the values of $R$ and $\theta.$

Part (a)

We know that the arc length is 25 cm.
. . So we have: . $R\theta\:=\:25$ [1]

We know that the area of the sector is 72 cm².
. . So we have: . $\frac{1}{2}R^2\theta\:=\:72$ [2]

Part (b)

Divide [2] by [1]: . $\frac{\frac{1}{2}R^2\theta}{R\theta} \:=\:\frac{72}{25}\quad\Rightarrow\quad\boxed{R \,=\,\frac{144}{25}\text{ cm}}$

Substitute into [1]: . $\frac{144}{25}\theta\:=\:25\quad\Rightarrow\quad\b oxed{\theta \,=\,\frac{625}{144}\text{ radians}}$