Answer:
x = 28°
Explanation:
PQ is tangent to circle O, ∠OPQ = 90°
∠QPR = 90-59 = 31°
mOP = mOR (radius) ∠PRO = ∠OPR = 59° (isosceles triangle)
∠PRO = ∠QPR + x° (triangle exterior angle)
x = 59° - 31° = 28°
28
Neat Question.
Thanks for posting.
The straign line (horizontal) makes an isosceles triangle inside the circle. So the bottom angle of the triangle is also 59 degrees. The central angle of the interior triangle is
180 - 59 - 59 = 62.
x and the 62 degree angle are complementary -- both are in the 90 degree angle triangle made by the tangent. x + 62 = 90
x = 90 - 62 = 28
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