Question

In any triangle abc the value of a(b^2+c^2)cos A+b (c^2+a^2)cos B+c(a^2+b^2)cos C is

Answer 1

c^2 = a^2 + b^2 - 2ab cos(C)

b^2 = a^2 + c^2 - 2ac cos(B)

a^2 = b^2 + c^2 - 2bc cos(A)

Answer 2

cos A = (b² +c² -a²)/2bc
cos B = (a² +c² -b²)/2ca
cos C = (a² +b² -c²)/2ab

Answer 3

c^2+2abcosc=a^2+b^2

Answer 4

b^2+2ac cos(B)=a^2 + c^2

Answer 5

a^2 +2bc cos(A) = b^2 + c^2

Answer 6

a(a^2+2bc cos(a))cos a is it correct? plug b^2+c^2 value

Answer 7

i am getting \(3abc\) as final answer after lot of algebra, there must be some simpler way >.<

Answer 8

yes @ganeshie8 3abc is the correct answer

Answer 9

u got it xD;)

Answer 10

yes but it took me two pages :o

Answer 11

I have used these :
cos A = (b² +c² -a²)/2bc
cos B = (a² +c² -b²)/2ca
cos C = (a² +b² -c²)/2ab

Answer 12

oh show the computation

Answer 13

heyy i got an idea whether it is right or not i dont know

Answer 14

a/sin A = b/sin B = c/sin C

Answer 15

oh sorry its sine rule

Answer 16

\[
\begin{array} \\
&= a(b^2+c^2)\cos A+b (c^2+a^2)\cos B+c(a^2+b^2)\cos C \\

&= a(b^2+c^2)\dfrac{b^2+c^2-a^2}{2bc}+b (c^2+a^2)\dfrac{c^2+a^2-b^2}{2ca}+c(a^2+b^2)\dfrac{a^2+b^2-c^2}{2ab} \\


&= a^2(b^2+c^2)\dfrac{b^2+c^2-a^2}{2abc}+b^2 (c^2+a^2)\dfrac{c^2+a^2-b^2}{2abc}+c^2(a^2+b^2)\dfrac{a^2+b^2-c^2}{2abc} \\

\end{array}

\]

Answer 17

i thought that but i wont get answer if its lengthy

Answer 18

u great done this

Answer 19

\[
\begin{array} \\


&= \dfrac{1}{2abc} \left( a^2(b^2+c^2)(b^2+c^2-a^2)\color{red}{+}b^2 (c^2+a^2)(c^2+a^2-b^2)\color{red}{+}c^2(a^2+b^2)(a^2+b^2-c^2)
\right) \\
\end{array}

\]

Answer 20

im sure there is a better way to do this -.-

Answer 21

yes

Answer 22

(a/cosA)=(b/cosB)=(c/cosC)

Answer 23

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

Answer 24

a^2+b^2=c^2

Answer 25

\[a^3 \cos A+b^3 \cos B+c^3 \cos C\]