if the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O such that angle C + angle D equals to ke angle AOB, then find the value of k.
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good time for a while since I've been able too much
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First of all, you have to know these two rules:
∠A + ∠B + ∠C + ∠D = 360 (For four angles and four sides)
∠A + ∠B + ∠AOB = 180 (for triangles)
∠AOB = 180 - [(∠A +∠B)/2]
∠C + ∠D = k. ∠AOB, where ∠AOB = 180 - [(∠A +∠B)/2]
So, ∠C + ∠D = k. [(360 - [ ∠A + ∠B] )/2]
2[∠C + ∠D]= k. (360 - [ ∠A + ∠B] ), where ∠C + ∠D = 360- (∠A + ∠B)
2[360-(∠A + ∠B)] = k. (360 - [ ∠A + ∠B])
By dividing both sides by (360 - [ ∠A + ∠B]) we will get:
2=k or k=2.
Hope it is helpful!
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