Question

If origin is the orthocentre of a triangle formed by the points $(cos\alpha ,sin\alpha ,0),(cos\beta ,sin\beta ,0),(cos\gamma ,sin\gamma ,0)$ then $\sum cos(2\alpha -\beta -\gamma )=$

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

As origin H(0,0,0) is the orthocenter of triangle formed by given point

As OA$\perp$BC

Slope Of OA$\times$slope of BC=-1

$\therefore \frac{sin\alpha }{cos\alpha }\times \frac{sin\gamma -sin\beta }{cos\gamma -cos\beta }=-1$

$\to tan\alpha .\frac{2cos\frac{\gamma +\beta }{2}sin\frac{\gamma -\beta }{2}}{-2sin\frac{\gamma +\beta }{2}sin\frac{\gamma -\beta }{2}}=-1$

$\to tan\alpha =tan\frac{\gamma +\beta }{2}$

$\to 2\alpha =\beta +\gamma$

Similarly, OB$\perp$AC

so we get $2\beta =\alpha +\gamma$

OC$\perp$AB

$\to 2\gamma =\alpha +\beta$

So $\sum cos(2\alpha -\beta -\gamma )=cos\left(0\right)=1$