In quadrilateral $$ABCD$$, it is given that $$AD = 8$$, $$DC = 12$$, $$CB = 10$$, and $$\angle A = \angle B = 60^\circ$$. Find the length of $$AB$$. Generalize.

Note by Ahaan Rungta
3 years, 9 months ago

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Let $$AD$$ and $$BC$$ intersect at $$E.$$ Notice that $$\triangle EAB$$ is equilateral. If $$AB=x,$$ we have $$CE = x-10$$ and $$DE= x-8$$. By Cosine rule on $$\triangle ECD$$, $12^2 = (x-8)^2 + (x-10)^2 - 2 \cdot \dfrac{1}{2} \cdot (x-8)(x-10) .$ Solving yields $$x= 9+\sqrt{141}$$.

The generalization attempt should work out similarly-- denote $$AD \cap BC = X,$$ use sine rule to find the ratios $$\dfrac{CX}{XB}$$ and $$\dfrac{DX}{AX},$$ and then use cosine rule on $$\triangle XAB.$$ I'm too lazy to carry out the details at the moment.

- 3 years, 9 months ago

Correct; nice solution!

- 3 years, 8 months ago