In a `DeltaABC`, AD is the altitude from A. Given `b gt c, angleC=23^(@)" and "AD=(abc)/((b^(2)-c^(2))`, find `angleB`.

Options

A. `83^(@)`
B. `97^(@)`
C. `113^(@)`
D. `127^(@)`

Answer

C

Explanation

`Delta = (1)/(2) a xx AD`
`rArr AD = (2Delta)/(a) = (abc)/(b^(2) -c^(2))` (given)
`rArr (2Delta)/(a) = (4R.Delta)/(b^(2) -c^(2))`
`rArr 2Ra = b^(2) -c^(2)`
`rArr sin A = sin^(2) B - sin^(2) C` (Using Sine rule)
`= sin (B + C) sin (B -C)`
`= sin A sin (B -C)`
`rArr sin(B -C) = 1`
`rArr angle B - angle C = (pi)/(2)`
`rArr angle B = (pi)/(2) + angle C = 90^(2) + 23^(@) = 113^(2)`