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P(cosalpha,sinalpha), Q(cosbeta, sinbeta...

`P(cosalpha,sinalpha), Q(cosbeta, sinbeta) , R(cosgamma, singamma)` are vertices of triangle whose orthocenter is `(0, 0)` then the value of `cos(alpha-beta) + cos(beta-gamma) + cos(gamma-alpha)` is

A

`-3//2`

B

`-1//2`

C

`½`

D

`3//2`

Text Solution

Verified by Experts

The correct Answer is:
A
Clearly `OP = OQ = OR`, where O is orthocentre
`:. O(0,0)` is the circum centre of `DeltaPQR`.
Also, given orthocenter `= (0,0)`
Thus, orthocentre and circumcentre coincide, So, triangle is equilateral
`:.` Centroid of `DeltaPQR = (0,0)`
`rArr cos alpha +cos beta +cos gamma =0`, and `sin alpha + sin beta+ sin gamma = 0` Squaring and adding we get
`cos(alpha-beta) +cos (beta-gamma) +cos (gamma- alpha) =-(3)/(2)`
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