1.

The angles of a triangle `A B C`are in `AdotPdot`and it is being given that `b : c=sqrt(3):sqrt(2)`, find `/_Adot`

Answer» As angles are in A.P.,
`:. 2/_B = /_A+/_C`
Now, `/_A+/_B+/_C = 180`
`/_B+2/_B = 180`
`/_B = 60^@`
Now, from sine law,
`b/sinB = c/sinC`
`b/c = sinB/sinC`
`=>sqrt3/sqrt2 = sin60^@/sinC`
`sinC = sqrt3/2*sqrt2/sqrt3 = 1/sqrt2`
`:. C = 45^@`
`/_A = 180-(60+45) = 75^@`


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