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```
if A=(cosa,sina,0),B=(cosb,sinb,0),C=(cosc,sinc,0) are vertices of triangle abc and cosa+cosb+cosc=3a,sina+sinb+sinc=3b,then orthocentre is 1.(3a,3b,0) 2.(2a,2b,0) 3.(0,0,0) 4.(a,b,0)
if A=(cosa,sina,0),B=(cosb,sinb,0),C=(cosc,sinc,0) are vertices of triangle abc and cosa+cosb+cosc=3a,sina+sinb+sinc=3b,then orthocentre is1.(3a,3b,0)2.(2a,2b,0)3.(0,0,0)4.(a,b,0)

```
9 months ago

```							dear student, this is a very ez ques. note that distance of all points A, B and C from the origin (0,0,0) is 1 bcoz sin^2theta+cos^2theta= 1. this means that A, B, C lie on the unit circle x^2+y^2=1, z= 0. Since the centre of this circle is (0,0,0), we conclude that the circumcentre of tri ABC is (0,0,0).now, we know that euler line passes through orthocentre (H), centroid (G) and circumcentre (O) and divides G in the ratio 2:1. also, g= (cosa*i + sina*j + 0*k + cosb*i + sinb*j + 0*k + cosc*i + sinc*j + 0*k)/3= [(cosa+cosb+cosc)i + (sina+sinb+sinc)j + 0k]/3or g= ai + bj so, G= (a, b, 0).apply section formula:assume position vector of H as xi + yj + zkai + bj = (2(0i + 0j + 0k) + 1(xi + yj + zk))/3or xi + yj + zk= 3ai + 3bjso, x= 3a, y= 3b and z=0so, H is (3a,3b,0)KINDLY APPROVE :))
```
9 months ago
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