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

Explanation

`cos B = (a^(2)+c^(2)-b^(2))/(2ac)`
`=a/(2c) - (b^(2)-c^(2))/(2ac)`
`a/(2c) -b/(2(AD))`
`=(sinA)/(2sinC) -1/(2sinC)` In `triangleACD`,
`cosB=(sinA-1)/(2sinC)` `(AD)/b= sinC rArr b/(AB) = 1/(sinC)`
`rArr 2cosB sinC = sinA-1`
`rArr sin(B+C)-sin(B-C)=sinA-1`
`rArr sinA-sin(B-C)=sinA-1`
`rArr sin(B-C)=1`
`(B-C) = pi/2`
`B=C+pi/2=23+90=113^(@)`
`angleB=113^(@)`