An isosceles triangle ABC is inscribed in the circle whose equation is `x^2 + y^2`=9 with vertex at A(3, 0) with base angles B and C each equal to 75º, then the product of the ordinates of B and C is :
| A. |
`-9/4` |
 |
| B. |
`9/4` |
| C. |
`3/4` |
| D. |
1 |
SInce` /_B = /_C`=` 75º
`/_BAC = 30º, /_BOC = 60º`
`/_OBC` is equilateral with BC = OB = 3
M is the mid point of BC.
OM=`sqrt(9-9/4)=(3sqrt3)/2`
Solving with `x^2+y^2 =9`, we get the points `(-(3sqrt3)/2,+-3/2)`
Therefore, product of the ordinates of B and C
`=3/2(-3/2)=-9/4`