1.

If `sinA=sin^2Ba n d2cos^2A=3cos^2B`then the triangle `A B C`isright angled (b) obtuse angled(c)isosceles (d) equilateral

Answer» `sinA = sin^2B ->(1)`
`2cos^2A = 3cos^2B`
`=>2cos^2A = 3(1-sin^2B)`
From (1),
`=>2(1-sin^2A) = 3(1-sinA)`
`=>2sin^2A - 3sinA+1 = 0`
`=>2sin^2A - 2sinA-sinA+1 = 0`
`=>2sinA(sinA-1)-1(sinA-1) = 0`
`=>(2sinA-1)(sinA-1) = 0`
`=>sinA = 1 or sinA = 1/2`
`=>A = 90^@ or A = 30^@`
When `A = 90^@`, `sin^2B = 1 => B = 90^@`, which is not possible for a triangle.
When `A = 30^@`, `sin^2B = 1/2 => B = 45^@`
`=>C = (180-30-45)^@ = 75^@`
`:. Delta ABC` is an obtuse angle triangle.
So, option-`(b)` is the correct option.


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