# There is a ABC triangle. (b+c)/(2k-1)=(c+a)/(2k)=(a+b)/(2k+1) is given where k>2 but not equal to 4. k is an integer. Show sinA/(k+1)=sinB/k=sinC/(k-1)

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### 1 Answer

By the law of the sines, `sinA/a = sinB/b = sinC/c` , hence, you should write a,b,c in terms of k, using the given relation such that:

`(b+c)/(2k-1)=(c+a)/(2k) `

`2k(b+c) = (2k-1)(c+a)`

Opening the brackets yields:

`2kb + 2kc = 2kc + 2ka - c - a =gt 2kb = 2ka - c - a `

`(b+c)/(2k-1)=(a+b)/(2k+1) =gt (2k+1)(b+c) = (2k-1)(a+b)`

Opening the brackets yields:

`2kb + 2kc + b + c = 2ka + 2kb - a - b`

`2kc + 2b = 2ka - a - c`

Notice that you may substitute `2kb` for `2ka - a - c` such that:

`2kc + 2b = 2kb =gt kc = b(k-1) =gt b/c = k/(k-1)`

You need to consider the given relation such that:

`sinB/k=sinC/(k-1) =gt ksinC = (k-1)sinB=gt k/(k-1) = sinB/sin C`

Considering the law of sines yields:

`sinB/b = sinC/c =gt b sinC = c sinB =gt b/c = sinB/sin C`

But `b/c = k/(k-1) =gt sinB/sin C = k/(k-1) =gt sinB/k=sinC/(k-1)`

**Hence, using the law of sines and the relation `(b+c)/(2k-1)=(c+a)/(2k)=(a+b)/(2k+1)` that helps to write a,b,c in terms of k yields `sinA/(k+1)=sinB/k=sinC/(k-1).` **

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