In a `Delta ` ABC the tangent of half the difference of two angles is one thired the tangement of half the sum of the angles .Determine the ratio of the sides opposite to the angles .

Explanation

Here, `tan (A-B)/(2)= 1/3 tan (A+B)/2 …..(i)`
using Napier's analogy `tan(A-B)/(2)= (a-b)/(a+b) cot (C/2) …(ii) `
form (i) & (ii)
`1/3 tan ((A+B)/(2))=(a-b)/(a+b).cot (C/2) rArr 1/3 cot (C/2)= (a-b)/(a+b)cot (C/2)`
`{as A+B+C = pi therefore tan ((B+C)/2)= tan (pi/2-C/2)=cot C/2}`
`rArr (a-b)/(a+b) or 3a-3b=a+b`
` 2a=4b or a/b=2/1rArr b/a = 1/2 `
Thus the ratio of the sides opposite to the angles is : a= 1:2