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in a triangle ABC, AD i the altitude from A givven b>c, angle C =2 degree and AD= (abc)/(b^2-c^2) find angle b
AD=c sinB=abc/(b² - c²) or sinB=ab/(b² - c²) or sinB=sinA sinB/sin(B+C) sin(B-C) or sinB=sinAsinB/sinAsin(B-C) or sin(B-C)=1=sin90° or B-C=90°but C = 2so angle B = 92 degree
AD=c sinB=abc/(b² - c²)
or sinB=ab/(b² - c²)
or sinB=sinA sinB/sin(B+C) sin(B-C)
or sinB=sinAsinB/sinAsin(B-C)
or sin(B-C)=1=sin90°
or B-C=90°
but C = 2
so angle B = 92 degree
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