Question

a (b cos C – c cos B) = b^2 – c^2

Answer 1

@radar

Answer 2

what is your question?

Answer 3

proving with law of sines and cosines

Answer 4

well in triangle ABC we have \[b^2=a^2+c^2-2ac \cos B\]\[c^2=a^2+b^2-2ab \cos C\]now evaluate \(a (b \cos C – c \cos B)\) and see if that is equal to the right hand side of given equation

Answer 5

when bracket is opened and multiplied and divided by 2

Answer 6

we get part of it

Answer 7

u can substitute \(\cos B\) and \(\cos C\) from equations i wrote using law of cosins

Answer 8

oh so let me check...

Answer 9

got it

Answer 10

\[a\left (b \frac{a^2+b^2-c^2}{2ab}-c \frac{a^2+c^2-b^2}{2ac} \right)=b^2-c^2\]

Answer 11

ok

Answer 12

i was stuck half way. when you gave the formula i got it

Answer 13

when bracket is opened and multiplied by 2 we get it

Answer 14

thanks

Answer 15

welcome