Circles !!

- anonymous

Circles !!

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- anonymous

|dw:1332961630235:dw|

- anonymous

how do i find x?

- anonymous

What is \[x^2-5x?\]

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## More answers

- anonymous

its the angel of ARC PQ

- anonymous

angle

- anonymous

what do you mean

- experimentX

what is 4x

- anonymous

its the angle od CQR sorry i forgot to label C . C is the middle

- anonymous

angle of*

- anonymous

Answer is 13 degrees

- experimentX

solve this
2(4x) = x^2−5

- anonymous

how did u get 13? and why are you multiplying 2 by 4x?

- anonymous

QCR=180-PCQ and also QRC=4x.So add all anges to get 180.
180-(x^2-5x)+8x=180.Solve this to get angle as 13 degrees

- anonymous

QRC is 4x because it is an isosceles triangle i.e.QC=CR

- anonymous

how do you know its an isosceles?

- anonymous

QC=CR=Radius of the circle

- experimentX

best of luck people

- anonymous

oh!!!

- anonymous

ok hold on let me solve it

- anonymous

what about QR? that little part left because of the triangle. that doesnt matteR?

- anonymous

where QR is i mean

- anonymous

& why is it 8x? shouldnt it be 12x? cuhs there are 3 angles in the triangle??

- anonymous

That doesn't matter all we need to figure out is that the triangle is isosceles hence QCR is 4x because what we want are the angles.The other side if you want you can find out using the cosine rule if you want but you don't need it here

- anonymous

oh ok.

- anonymous

can you help me with another problem?

- anonymous

Sure

- anonymous

A clock with hour and minute hands is set to 1:00 PM. After 20 minutes, what will be the measure of the minor arc formed by the hour and minute hands? I have no clue what to do

- anonymous

Ok let me see

- anonymous

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- anonymous

im sorry but i have no clue what that means or what you did.

- anonymous

Now the clock is divided into 12 parts.Total angle is \[2\pi\] radians so angle contained in each part is \[2\pi/12\]which is \[\pi/6\] Now here for the theta we need the arc encloses 3 parts so theta is (pi/6)*3

- anonymous

Length of the arc is radius*angle enclosed by arc

- anonymous

So now theta is pi/2 so length of arc=r*pi/2

- anonymous

where did you get pi from

- anonymous

Total angle in a circle is \[2\pi radians\]

- anonymous

Did you understand?

- anonymous

uhmm hold on. im looking at what you did

- anonymous

btw whats theta?

- anonymous

The angle enclosed by the arc

- anonymous

\[\Theta \] is theta

- anonymous

doesnt it enclose 4 parts?

- anonymous

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- anonymous

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- anonymous

isnt that after 20 mins?

- anonymous

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- anonymous

The time is 1.20

- anonymous

Coincidentially the time here where I live is 1.20!Lol

- anonymous

OOHH ok

- anonymous

so can you explain what you did again ? i dont get how you got pi and R . did u use the circumference? C=2pi(r)?

- anonymous

No now the whole angle in the circle is 2 pi

- anonymous

And the arc encloses 3/12 th of that

- anonymous

how is the whole angle 2 PI?

- anonymous

360 degrees=\[2\pi\] radians.Radian is another way of measuring an angle.
http://en.wikipedia.org/wiki/Radian

- anonymous

ohhhhhh. okay. but if i was to do everything you did & used 360 instead of 2 pi would it be wrong?

- anonymous

Yeah it would.The formula is valid only if the angle is measured in radians.

- anonymous

so, is the answer pi/2?

- anonymous

For example the whole angle inside a circle so length of arc by circle would be 2pi*r which is actually the circumference

- anonymous

No the angle enclosed is pi/2 the length of the arc is r*pi/2

- anonymous

& what is r? Radian?

- anonymous

No r is the radius.The formula is length of arc is the multiple of angle enclosed in radians and radius of the arc

- anonymous

http://en.wikipedia.org/wiki/Arc_(geometry)#Arc_length

- anonymous

oh ok . i think i get it now.
thanks!

- anonymous

hey can you help me with 1 more? i cant solve it.

- anonymous

Sure go on

- anonymous

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- anonymous

The plane intersects the sphere in a circle that has diameter 12. If the diameter of the sphere is 18, what is the value of x? Give ur answer in simplified radical form.

- anonymous

Ok sounds interesting.Let's see

- anonymous

I am not sure how to do this.Ill tell you if I get it

- anonymous

ok

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