In acute angled triangle `A B C ,A D` is the altitude. Circle drawn with `A D` as its diameter cuts `A Ba n dA Ca tPa n dQ ,` respectively. Length of `P Q` is equal to `/(2R)` (b) `(a b c)/(4R^2)` (c)`2RsinAsinBsinC` (d) `delta/R`

Options

A. `(Delta)/(2R)`
B. `(abc)/(4R^(2))`
C. `2R sin A sin B sin C`
D. `(Delta)/(R)`

Answer

C, D

Explanation

`AD = c sin B`
Let R' be circumradius of `DeltaAPQ`.
Using sine rule in triangle APQ, we get
`(PQ)/(sin A) = 2R' = AD = c sin B`
`rArr PQ = c sin B sin A = c (b)/(2R) sin A = (Delta)/(R)`
Also, PQ `= c sin B sin A`
`=2R sin C sin B sin A`