ABC एक चक्रीय त्रिभुज है और `angleBAC` `angleABC`और `∠BCA` के समद्विभाजक क्रमश : P , Q तथा R बिंदु वृत्त पर मिलते है , तो `angleRQP=` ?

Options

A. `90^(@)-B/2`
B. `90^(@)-C/2`
C. `90^(@)-A/2`
D. `90+B/2`

Answer

A

Explanation

a)
From chord BP
`angleBAP = angleBQP` = z(Angle substended by same chord on circumference)
Similarly from chord RB
`angleRCB=angleRQB = y`
In `DeltaABC`
`2x + 2y + 2z = 180^(@)`
x+y+z = `90^(@)`
y+z=`90^(@)-x`
`angleRQP = 90^(@)-x`
`angleRQP = 90^(@)-angleB/2`