Question

the law of cosines is a2+b2-2abcosc=c^2 find the value of 2abcosc

Answer 1

For this case we have the following equation:

$a ^ 2 + b ^ 2-2abcosC = c ^ 2$

To find the value of 2abcosC we must follow the following steps:

Subtract a ^ 2 on both sides of the equation:

$-a ^ 2 + a ^ 2 + b ^ 2-2abcosC = c ^ 2-a ^ 2 b ^ 2-2abcosC = c ^ 2-a ^ 2$

Subtract b ^ 2 on both sides of the equation:

$-b ^ 2 + b ^ 2-2abcosC = c ^ 2-a ^ 2-b ^ 2 -2abcosC = c ^ 2-a ^ 2-b ^ 2$

Multiply both sides of the equation by -1:

$(-2abcosC) (- 1) = (c ^ 2-a ^ 2-b ^ 2) (- 1) 2abcosC = -c ^ 2 + a ^ 2 + b ^ 2$

Answer:

the value of 2abcosC is:

$2abcosC = -c ^ 2 + a ^ 2 + b ^ 2$

Answer 2

Answer:

The value of  $2ab\cos C$ is  $a^2+b^2-c^2$

Step-by-step explanation:

 Given : The laws of cosine is $a^2+b^2-2ab\cos C=c^2$

We have to find the value of  $2ab\cos C$

Consider the given formula for laws of cosine,

$a^2+b^2-2ab\cos C=c^2$

Subtract both side $c^2$, we have,

$a^2+b^2-2ab\cos C-c^2=0$

Add  $2ab\cos C$ both side, we have,

$a^2+b^2-c^2=2ab\cos C$

Thus,  The value of  $2ab\cos C$ is  $a^2+b^2-c^2$