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?

Question

1370 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-26T14:27:26

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#