In the ambiguous case if the remaining angles of a triangle with given a, b, A and `B_(1),B_(2),C_(1),C_(2)` then `(sin C_(1))/(sin B_(1))+(sin C_(2))/(sin B_(2))=`

Options

A. 2 cos A
B. 2 sin B
C. 2 tan A
D. 2 cot A

Answer

A

Explanation

`a^(2)=b^(2)+c^(2)-2bc` coa A
or `c^(2)-(2b cos A)c+b^(2)-a^(2)=0`
Above equation has two roots `c_(1)` and `c_(2)`
`therefore c_(1)+c_(2)=2bcos A` and `c_(1)c_(2)=b^(2)-a^(2)`
`sin B_(1)=sin B_(2)=(b sin A)/(a)`
`sin C_(1)=(c_(1)sin A)/(a)`
`sin C_(2)=(c_(2)sin A)/(a)`
`therefore (sin C_(1))/(sin B_(1))+(sin C_(2))/(sin B_(2))=(c_(1)+c_(2))/(b)=2 cos A`