There exists a triangle `A B C` satisfying the conditions `bsinA=a ,A a ,A >pi/2` `bsinA > a ,A>a` `bsinA<>pi/2,b=a`

Options

A. `tan A + tan B + tan C = 0`
B. `(sin A)/(2) = (sin B)/(3) = (sin C)/(7)`
C. `(a + b)^(2) = c^(2) + ab and sqrt2 (sinA + cos A) = sqrt3`
D. `sin A + sin B = (sqrt3 + 1)/(2), cos A cos B = (sqrt3)/(4) = sin A sin B`

Answer

C, D

Explanation

(1) `tan A + tan B + tan C = tan A tan B tan C = 0`
Therefore, either A or B or C = 0. This is not possible in a triangle
(2) By sine rule, `(a)/(2) = (b)/(3) = (c)/(7) = lamda`(say)
`a + b = 5 lamda, c = 7 lamda` i.e., `a + b lt c`
This is not possible in a triangle, as the sum of two sides is greater than the third.
(3) Given that `(a + b)^(2) = c^(2) + ab`
or `a^(2) + b^(2) - c^(2) + ab = 0`
or `2ab cos C + ab = 0`
or `cos C = -(1)/(2) or angleC = 120^(@)`
Also `sqrt2 (sin A + cos A) = sqrt3`(given)
or `sin (A + 45^(@)) = (sqrt3)/(2)`
or `A = 45^(@) = 60^(@) " " (A + 45^(@) = 120^(@) " out as " angle C )`
or `A = 15^(@) and "hence " B = 45^(@)`
`:. Delta` is possible
(4) Given that `cos a cos B = (sqrt3)/(4) = sin A sin B`
`rArr cos (A -B) = (sqrt3)/(2)`
or `A - B = (pi)/(6)`
and `cos (A +B) = 0`
or `A +B = (pi)/(2)`
`rArr A = 60^(@), B = 30^(@)`
Hence, `sin A + sin B = (sqrt3+1)/(2)`. This is possible in a triangle