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the diameter CD could opposite the angle 90` if the chords CB and AD crossed not inside the circle but their cross point was on the circle. Here we may not assume that this is the case, because the cross point is not on the circle, it's in the circle.tomada wrote:Maybe I'm looking at this too simplistically but, if CD is a diameter of the circle AND a side of a triangle (which triangle, please specify) inscribed within that circle, doesn't the angle opposite that side = 90 degrees? This would mean that x = 180 - 25 - 90 = 65 degrees, and we wouldn't need to know the length of the arc AB.
if CD is a diameter can x be more than 90`?tomada wrote:In any case, it's interesting that, thus far, two very different values of 'x' have been calculated - 135 and 45.
this is only true if the triangle is indeed inscribed = all three vertices on the circumference, which isn't true in this case.tomada wrote:Maybe I'm looking at this too simplistically but, if CD is a diameter of the circle AND a side of a triangle inscribed within that circle, doesn't the angle opposite that side = 90 degrees? This would mean that x = 180 - 25 - 90 = 65 degrees, and we wouldn't need to know the length of the arc AB.
yellowho wrote:Geva,
1) Just to be clear when you said central angle: Assuming the circle has center point O, central angle of angle C = DOB?
Indeed.
2) Segment CB and AD intersect. Lets call that point Z. Is there a way to tell what angle Z is from knowing the minor arc AB? (I know you can do 180-25+45=110 way but just wondering if there's another way. For some reason I have in my head that theres a property where angle AZB is half of AOB. I was trying to apply that knowledge.)
No easy way to find the angle AZB. A circle recognizes only two "special" types of angles - central angles (two points on the circumference, one point on the center) and inscribed angles (all three points on the circumference). Angle ZAB falls into neither of these categories.
Note that angle AZB can't be half of AOB - if you draw the central angle, you can see visually that AOB will be smaller than AZB, so it can't logically be twice as great as AZB. You're confusing central angles and inscribed angles, which do follow this property.
Geva@MasterGMAT wrote:circumference is 36, so semicircle's circum is half = 18.
this circumference is composed of three arcs
arc defined by 25 degree inscribed angle
arc defined by inscribed angle x
arc ab, given as 4.
recall that an inscribed angle is half of the central angle lying on the same arc.
so the 25 angle corresponds to a 50 degree central angle, which corresponds to arc length 50/360*36 = 5.
so arc length of x is 18-5-4 = 9.
meaning hat central angle of x is 9/36*360 = 90, and x is half of that - 45.
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