Question

in a triangle ABC, a[bcos C-c cos B]=​

Answer 1

Step-by-step explanation:

cosC=(a2+b2-c2)÷2ab

cosB=(c2+a2-b2)÷2ac

so bcosC=(a2+b2-c2)÷2a

and ccosB=(c2+a2-b2)÷2a

then a[bcosC-ccosB]={(a2+b2-c2)÷2}-{(c2+a2-b2)÷2}

=(b2-c2)

In the triangle ABC