Question

Law of Cosines

HELP!!!!!

Answer 1

`\displaystyle 21 \approx c`

Step-by-step explanation:

Solving for Angles

`\displaystyle (a^2 + b^2 - c^2)/(2ab) = cos\angle{C} \\ (a^2 - b^2 + c^2)/(2ac) = cos\angle{B} \\ (-a^2 + b^2 + c^2)/(2bc) = cos\angle{A}`

Do not forget to use
`\displaystyle arccos` or
`\displaystyle cos^(-1)`towards the end, or the result will be thrown off.

Solving for Edges

`\displaystyle b^2 + a^2 - 2ba\:cos\angle{C} = c^2 \\ c^2 + a^2 - 2ca\:cos\angle{B} = b^2 \\ c^2 + b^2 - 2cb\:cos\angle{A} = a^2`

Take the square root of the result in the end, or you will throw yourself off.

Well, let us get to work:

`\displaystyle 13^2 + 29^2 - 2[13][29]cos\:41 = c^2 \\ 169 + 841 - 754cos\:41 = c^2 \\ 1010 - 754cos\:41 = c^2 \\ \\ √(440,94897651...) = √(c^2) \\ 20,99878512... = c \\ \\ \boxed{21 \approx c}`

I am joyous to assist you at any time.