Posted by lilman on Tuesday, May 22, 2012 at 7:04pm.
According to the wording of part a) your description of
QP= 8x-10, PR = 6x etc
must have said
arc QP = 8x-10 etc
so 8x-10 + 6x + 10x+10 = 360
24x = 360
x = 15
so arc QP = 8(15) - 10 = 110°
PR = 90°
QR = 160°
By the inscribed angle theorem, the angle opposite the arc must be half the central angle subtended by the arc. Then....
angle R = 55°
angle Q =45°
angle P = 80°
looks like plain old scalene triangle
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