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.

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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. – Sreejato Bhattacharya · 2 years, 1 month ago

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– Ahaan Rungta · 2 years, 1 month ago

Correct; nice solution!Log in to reply