1.

Line OQ is angle bisector of angle O of right angle triangle OPR, right angle at P. Point Q is such that ORQP is concyclic. If point O is orign and points P,Q,R are represented by the complexnumbers z_(3),z_(2),z_(1) respectively. If (z_(2)^(2))/(z_(1)z_(2))=3/2 then (R is circum radius of /_\OPR)

Answer»

Angles of `/_\OPR` are `(pi)/6,(pi)/3,(pi)/2`
Angles of `/_\OPR` are `(pi)/4,(pi)/4,(pi)/2`
Area of `/_\OPR` is `2sqrt(2)R^(2)`
Area of `/_\OPR` is `2R^(2)`

Solution :By Rotation theory
`(z_(2))/(z_(1))=costhetae^(itheta)`
`(z_(3))/(z_(1))=(COS2THETA)/(costheta).e^(itheta)`
So `(z_(2)^(2))/(z_(3)z_(1))=(cos^(2)theta)/(cos2theta)=3/2`
`theta=(pi)/6`


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