NDA 1 2024 Maths, question 63

Topic: Analytical geometry

Q63

$ABC$ 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 ?

(a) $\frac{1}{3}$
(b) $\frac{1}{2}$ Correct answer
(c) $\frac{2}{3}$
(d) $2$

Solution.

Answer: (b) $\frac{1}{2}$

Direction vectors (1, 7) and (1, −1) give the acute vertex angle A with cos A = 3/5 and sin A = 4/5.
For the equal base angles θ, θ = (π − A)/2.
Hence cot θ = tan(A/2) = sin A/(1 + cos A) = 1/2.