If in a `DeltaABC`, the tangent of half the difference of two angles is one-third the tangent of half the sum of the angles. Then, the ratio of the sides opposite to the angles is

Options

A. `2 : 1`
B. `1 : 2`
C. `3 : 1`
D. `1 : 1`

Answer

A

Explanation

According to the question,
`tan((A-B)/(2))=(1)/(3)tan((A+B)/(2))" ... (i)"`
Using Napier's rule.
`tan.((A-B)/(2))=(a-b)/(a+b).cot((C)/(2))" ... (ii)"`
From Eqs. (i) and (ii), we get
`(1)/(3).tan((A+B)/(2))=(a-b)/(a+b).cot((C)/(2))`
`implies(1)/(3)cot.(C)/(2)=(a-b)/(a+b)cot.(C)/(2)`
`implies (a-b)/(a+b)=(1)/(3)implies(2a)/(-2b)=(4)/(-2)` [by componendo and dividendo rule]
`implies (a)/(b)=(2)/(1)`
`therefore a:b=2:1`