Circle Geometry Question

• May 27th 2007, 06:39 AM
Danielisew
Circle Geometry Question
I absolutely hate circle geometry. I normally THINK i have found out the answer, but it ALWAYS asks 'give a reason for your answer'. This really annoys me, as i never know how to phrase the answer - whether to state a rule i have used or just how i did it!

Anyway, i got a question here that i want someone to do for me as i cannot do it and find this really hard :(

http://img529.imageshack.us/img529/903/99197842qp0.jpg

a) Find the size of angle PQR

b) Find the size of angle PRQ

c) Find the size of angle POQ

Can someone please give this a shot?

Thanks :)
• May 27th 2007, 07:08 AM
chandrasekhar_mallisetty
solution for circles problem
angle in a semicircle is a rightangle hence angle PQR is a rightangle:)

Quote:

Originally Posted by Danielisew
I absolutely hate circle geometry. I normally THINK i have found out the answer, but it ALWAYS asks 'give a reason for your answer'. This really annoys me, as i never know how to phrase the answer - whether to state a rule i have used or just how i did it!

Anyway, i got a question here that i want someone to do for me as i cannot do it and find this really hard :(

http://img529.imageshack.us/img529/903/99197842qp0.jpg

a) Find the size of angle PQR

b) Find the size of angle PRQ

c) Find the size of angle POQ

Can someone please give this a shot?

Thanks :)

• May 27th 2007, 07:13 AM
Danielisew
Well that answers part A, but i still don't understand your reason - wouldn't it be 'the angle in between two semi-circles is 90°' ?

Anyone do parts b and c please?
• May 27th 2007, 07:38 AM
Soroban
Hello, Danielisew!

Quote:

Wouldn't it be 'the angle in between two semi-circles is 90°' ?
I have no idea what this means . . .

You expected to know about inscribed angles:
. . An inscribed angle is measure by one-half its intercepted arc.

(a) Since PR is a diameter, angle PQR intercepts an arc of 180°.
. . .Therefore: . $\angle PQR = 90^o$

(b) $\angle PSQ = 56^o$ . . . Hence: . $\text{arc }PQ = 112^p$
. . .Angle PRQ intercepts $\text{arc }PQ = 112^o$
. . .Therefore: . $\angle PRQ = 56^o$

(c) Angle POQ is a central angle, measured by its intercepted arc.
. . .Its intercepted arc is: $\text{arc }PQ = 112^o$
. . .Therefore: . $\angle POQ = 112^o$

• May 27th 2007, 08:23 AM
chandrasekhar_mallisetty
solution for circles problem
Quote:

Originally Posted by Danielisew
Well that answers part A, but i still don't understand your reason - wouldn't it be 'the angle in between two semi-circles is 90°' ?

Anyone do parts b and c please?

angle in a semi circle is a right angle the reason is if QR is the diameter and P is on the circumference then find slope of QP multiplied with slope of RP gives -1 You can take center as origin, radius as a units then Q(a,0) and R(-a,0) and P(x,y)
Angle at the centre is double the at the circumference that gives answer for the 2nd question
Angles in the same segments are equal that gives the answer for the third.
• Jan 15th 2009, 11:16 AM
mathman101
answer to ur circle geometry question
a) will equal 45 degrees, because Q and R create a right angle triangle, which is divided by the line of SQ.
b) will be 56 degrees, because it is the same as angle S.
c) 92 degrees, because triangle POQ is equal to 180 degrees and is also an equalateral triangle, with angles P and Q equaling 44 degrees each, 180 degrees-88 degrees=92 degrees. i would ask around to confirm my answers, for this question stumbled me. however i am fairly certain of these answers.