Consider the following statements concerning a `DeltaABC`
(i) The sides a,b,c and area of triangle are rational.
(ii) `a, "tan"(B)/(2),"tan"(C)/(2)`
(iii) `a, sin A sin B, sin C` are rational .
Prove that `(i) rArr(ii) rArr(iii)rArr(i)`

Explanation

It is given that a, b, c and areao of triangle are rational .
We have `"tan"(B)/(2)=sqrt(((s-c)(s-a))/(s(s-b)))`
`=sqrt(((s-c)(s-a)(s-c))/(s(s-b)))`
Again, a, b c are rational give, `s=(a+b+c)/(2)` are rational Also, (s-b) is rational, since triangle is rational,, therefore, we get
`"tan"((B)/(2))=(Delta)/(s(s-b))` is rational
Similarly, `"tan"(C)/(2)=(Delta)/(s(s-c))` is rational.
Therefore, `a, "tan"(B)/(2),"tan"(C)/(2)` are rational.
which show that `(i) rarr(ii)`,
Again it given that
`a, "tan"(B)/(2),"tan"(C)/(2)` are rational, then
`"tan"(A)/(2)"tan" ((pi)/(2)-(B+C)/(2))`
`=cot ((B+C)/(2))=(1)/(tan((B)/(2)+(c)/(2)))`
`=(1-"tan"((B)/(2))*"tan"((C)/(2)))/("tan"((B)/(2))*"tan"((C)/(2)))`
Since tan (B/2) tan (C/2) are rational, hence tan (A/2) is a rational
Now, `sinA=(2 tan A//2)/(1+tan^(2)A//2)` as tan (A/2) is a rational number sin A is a rational. Similarly sin B and sin C are, thsu a sin A, sin B, sin C are rational, therefore `(ii)rArr(iii)`,
Again a, sin A, sin B, sin C are rational,
By the sin rule,
`(a)/(sinA)=(b)/(sinB)=(c)/(sinC)`
`rArr b=(asin B)/(sinA) and c=(a sin C)/(sin A)`
Since a , sin A , sin B and C are rational,
Hence, b and ca are also rational.
Also, `Delta=(1)/(2)bc sin A`
As b, c and sin A are rational, so triangle is rational number. therefore a,b,c and triangle are rationla.
Therefore, `(iii) rArr(i)`