If in triangle ABC, a, c and angle A are given and `c sin A lt a lt c`, then (`b_(1) and b_(2)` are values of b) (a) `b_(1) + b_(2) = 2c cos A` (b) `b_(1) + b_(2) = c cos A` (c)`b_(1) b_(2) = c^(2) -a^(2)` (d) `b_(1) b_(2) = c^(2) + a^(2)`

Options

A. `b_(1) + b_(2) = 2c cos A`
B. `b_(1) + b_(2) = c cos A`
C. `b_(1) b_(2) = c^(2) -a^(2)`
D. `b_(1) b_(2) = c^(2) + a^(2)`

Answer

A, C

Explanation

From the cosine formula, `cos A = (b^(2) + c^(2) -a^(2))/(2bc)`
or `b^(2) - (2c cos A) b + (c^(2) -a^(2)) = 0`,
Which is a quadratic equation in b. Therefore,
`c sin A lt a lt`
Therefore, two triangle will be obtained, But this is possible when two values of the thired side are also obtained. Clearly, two values of sides b will be `b_(1) and b_(2)`. Let these be the roots of the above equation. then,
Sum of roots `= b_(1) + b_(2) = 2c cos A and b_(1) b_(2) = c^(2) - a^(2)`