shuba.personal Wrote:I understood that angle RPQ has the same measurement as angle ORP (parallel lines), but got stuck at this point. Couldn't figure out how to create the isoceles triangle from here. Could someone explain? Thanks.
once you make that realization -- along with the realization that OPQR is a semicircle (which you know because they tell you that OR is a diameter) -- there are two ways you can go.
1/
the long way:
draw a radius from point P to the center of the circle. (let's call the center point X, since they haven't given it a name.) the triangle PXR thereby created is isosceles (because two of its sides are radii of the circle), so its angles are 35, 35, and 110.
this means that angle PXO is 180 - 110 = 70 degrees. that's a central angle, so the arc OP is the same (also 70 degrees).
now, ignore the diagonal line PR, and just concentrate on the left-right symmetry created by the two horizontal lines. you will see that the arc QR is the mirror image of the arc OP, so that arc is also 70 degrees.
finally, use the semicircle = 180 to figure out that arc PQ is 180 - 70 - 70 = 40 degrees. that's 40/360 = 1/9 of the whole circumference, giving you the answer.
2/
the short way:
there's a really obscure fact about "inscribed angles" (angles that have their vertex on the circle, like the angles QPR and PRO in this diagram): namely, the measure of the inscribed angle is exactly half the measure of the arc that it cuts off.
if you know this fact -- which, by the way, has never played a role in any other official problem, ever -- then you can cut right to the fact that the two arcs are 70 degrees each. from that point onward, the solution process is the same.
