Question

In a triangle ABC, if the tangent of half the difference of two angles is equal to one third of the tangent of half the sum of the angles, then the ratio of the sides opposite to the angles is

• A. 2:1
• B. 1:2
• C. 3:1
• D. 1:1

Hint:

Napier's Analogy (Law of Tangents) is a powerful tool for problems involving the sum and difference of angles and sides of a triangle. The key formula is \( \frac{a-b}{a+b} = \frac{\tan((A-B)/2)}{\tan((A+B)/2)} \). Recognizing this pattern immediately solves the problem.

Answer 1

The Correct Option is A

Let the two angles be A and B. We are given: \[ \tan\left(\frac{A-B}{2}\right) = \frac{1}{3} \tan\left(\frac{A+B}{2}\right) \] Rearranging this gives: \[ \frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{A+B}{2}\right)} = \frac{1}{3} \] This is a direct application of Napier's Analogy (or the Law of Tangents) in a triangle, which states: \[ \frac{\tan\left(\frac{A-B}{2}\right)}{\tan\left(\frac{A+B}{2}\right)} = \frac{a-b}{a+b} \] where a and b are the lengths of the sides opposite to angles A and B, respectively. Equating the two expressions: \[ \frac{a-b}{a+b} = \frac{1}{3} \] Cross-multiply to solve for the ratio of a to b: \[ 3(a-b) = 1(a+b) \] \[ 3a - 3b = a + b \] \[ 3a - a = b + 3b \] \[ 2a = 4b \] \[ \frac{a}{b} = \frac{4}{2} = \frac{2}{1} \] The ratio of the sides \( a:b \) is 2:1.