If `Delta = a^(2)-(b-c)^(2), Delta` is the area of the `Delta ABC` then `tan A =` ?

Options

A. `(8)/(14)`
B. `(8)/(13)`
C. `(8)/(15)`
D. `(8)/(17)`

Answer

C

Explanation

Given, `Delta=a^(2)-(b-c)^(2)=(a+b-c)(a-b+c)`
`implies Delta=[4(s-b)(s-c)]`
`impliess(s-a)(s-b)(s-c)=16(s-b)^(2)(s-c)^(2)`
`=((s-b)(s-c))/(s(s-a))=(1)/(16)`
`implies tan.(A)/(2)=(1)/(4)`
`therefore tanA=(2tan.(A)/(2))/(1-tan^(2).(A)/(2))=(2.((1)/(4)))/(1-(1)/(16))=(8)/(15)`