`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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CENGAGE ENGLISH-COORDINATE SYSTEM-Multiple Correct Answers Type