The area of triangle ABC is equal to a^2+b^2-c^2. If angle C is acute then find the numerical vaue of secant where a,b,c are positive real numbers?

1 Answer
Jul 26, 2018

Given Delta=a^2+b^2-c^2...(1)

We know

cosC=(a^2+b^2-c^2)/(2ab)....(2)

Delta=1/2 ab sinC

=>ab=(2Delta)/sin c....(3)

Combining these equtions we get

cosC=(DeltasinC)/(2*2Delta)

=>tanC=4

=>sec^2C=1+tan^2C=1+16=17

=>secC=sqrt17