Question

In ΔABC, a = 3, b = 1, cos(A – B) = `2/9`, find c.

Answer 1

We have

`tan((A - B)/2) = sqrt((1 - cos(A - B))/(1 + cos(A - B))`

= `sqrt((1 - 2/3)/(1 + 2/3))`

= `1/sqrt(5)`

Also, `tan((A - B)/2) = (a - b)/(a + b)*cot  C/2`   ...(Napier’s analogy)

∴ `1/sqrt(5) = (3 - 1)/(3 + 2)*cot  C/2`,

∴ `cot  C/2 = 2/sqrt(5)`

`\implies tan  C/2 = sqrt(5)/2`

But cos C = `(1 - tan^2  C/2)/(1 + tan^2  C/2)`

= `(1 - 5/4)/(1 + 5/4)`

= `-1/9`

∴ `(a^2 + b^2 - c^2)/(2ab) = -1/9`   ...(Cosine rule)

∴ `((3)^2 + (1)^2 - c^2)/(2(3)(1)) = -1/9` 

∴ –c² = `-32/3`

∴ c² = `32/3`,

`\implies` c = `(4sqrt(2))/sqrt(3)`