In `Delta ABC` it is given distance between the circumcentre (O) and orthocentre (H) is `R sqrt(1-8 cos A cos B cos C)`. If Q is the midopoint of OH, then AQ is

Options

A. `(R )/(2)sqrt(1+8cos A sin B sin C)`
B. `R sqrt(1+8cos A sin B sin C)`
C. `2R sqrt(1+8cos A sin B sin C)`
D. `(R )/(2)sqrt(1+8sin A cos B cos C)`

Answer

A

Explanation

We know that OA = R, HA = 2R cos A and
`OH = R sqrt(1-8cos A cos B cos C)`
Applying Apollonius theorem to `Delta AOH`, we get
`2(AQ)^(2)+2(OQ)^(2)=OA^(2)+(HA)^(2)`
`rArr 2(AQ)^(2)=R^(2)+4R^(2)cos^(2)A-(R^(2))/(2)(1-8cos A cos B cos C)`
`rArr 4AQ^(2)=2R^(2)+8R^(2)cos^(2)A-R^(2)+8 R^(2) cos A cos B cos C`
`rArr 4AQ^(2)=R^(2)(1+8cos^(2)A+8 cos A cos cos C)`
`=R^(2)(1+8cos A(cos A + cos B cos C))`
`=R^(2)(1+8cos A(cos B cos C - cos(B+C))`
`= R^(2)(1+8 cos A sin B sin C)`