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 θ is one of the equal angles, then what is cotθ equal to?

The angle A between the two equal sides lies between the lines with slopes 7 and –1. tan A = |(7 – (–1))/(1 + 7·(–1))| = |8/(–6)| = 4/3. Since the triangle is isosceles with AB = AC, the base angles θ are equal and 2θ + A = π, so θ = (π – A)/2. Then cotθ = cot((π – A)/2) = tan(A/2). Using tan A = 4/3, in 0 < A < π, sin A = 4/5, cos A = 3/5 (since cos A could be –3/5 too, but for acute triangle A is acute so cos A = 3/5). Then tan(A/2) = sin A/(1 + cos A) = (4/5)/(1 + 3/5) = (4/5)/(8/5) = 1/2. Hmm, that gives 1/2, option (b). Let me reconsider — for tan A = 4/3 with A acute, sin A = 4/5, cos A = 3/5, tan(A/2) = (1 – cos A)/sin A = (2/5)/(4/5) = 1/2. So cotθ = 1/2.