ABC is an acute angled isosceles triangle. Two equal sides AB and AC lie on the lines 7x-y-3=0 and x+y-5=0. If \theta is one of the equal angles, then what is \cot \theta equal to ?

Explanation

Let \angle A be the angle between the given lines. Their slopes are m_1=7 and m_2=-1. Then \tan A = |\frac{7 - (-1)}{1 + 7(-1)}| = |\frac{8}{-6}| = \frac{4}{3}. Since \triangle ABC is isosceles, \angle A + 2\theta = 180^{\circ}, so \tan 2\theta = \tan(180^{\circ}-A) = -\tan A = -\frac{4}{3}. Using \tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta} = -\frac{4}{3}, we get 2\tan^2\theta - 3\tan\theta - 2 = 0. Solving yields \tan\theta = 2 or -\frac{1}{2}. Since the triangle is acute-angled, \theta is acute, so \tan\theta = 2. Thus, \cot\theta = \frac{1}{2}.

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