O is the center of the circle; AB is the diameter

P is on OB.

the angle OQP = angle ORP = 10 degree

The angle AOQ = 40 degree

Find angle ROB

http://members.lycos.co.uk/ghostsingular/5.jpg

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- Mar 25th 2007, 09:54 AMSingularCircle
O is the center of the circle; AB is the diameter

P is on OB.

**the angle OQP = angle ORP = 10 degree**

**The angle AOQ = 40 degree**

Find angle ROB

http://members.lycos.co.uk/ghostsingular/5.jpg - Mar 27th 2007, 12:55 AMearboth
Hello,

maybe this comes too late, but nevertheless..

I only can offer you a more or less precise drawing with a result.

The thick black lines describe the situation of your problem. The thin blue lines are necessary to place Point R on the circle line. The red value is the result you are looking for.

Obviously the center of the blue circle is placed on the black circle too. Maybe this property is of some use for you.

EB

PS.: I corrected my drawing because I measured the wrong angle. - Mar 27th 2007, 06:21 AMearboth
Hello,

it's me again.

I've modified your sketch a little bit (see attachment):

OP is the base of the 2 triangles, therefore I called this line b.

The angle <(QPA) = α (alpha) and the angle <(RPA) = β (beta).

OQ = OR = r

You know that α + 10° = 40° thus α = 30°

Now use Sine rule to calculate b:

b/r = sin(10°)/sin(30°). Because sin(30°) = 0.5 you get:

b = 2rsin(10°)

Now use Sine rule to calculate β:

sin(β)/sin(10°) = r / b Solve for sin(β) and plug in the term for b:

sin(β) = (r sin(10°))/(2r sin(10°)) = 1/2

Therefore β = 30° or β = 150°. β must be greater than 90° therefore β = 150°. The angle <(ROB) = 180° - 150° - 10° = 20°

EB - Mar 30th 2007, 01:30 PMSingular
Thanks earboth :)