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

Evaluate

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

Swap the sides of the equation

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

Move the expression to the right-hand side and change its sign

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

If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses

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

Divide both sides

\frac{- 2 ab cos ( C )}{- 2 ab} = \frac{c ^{2} - a ^{2} - b ^{2}}{- 2 ab}

Divide the numbers

cos (C) = \frac{c ^{2} - a ^{2} - b ^{2}}{- 2 ab}

Divide the numbers

More Steps

cos (C) = \frac{- c ^{2} + a ^{2} + b ^{2}}{2 ab}

Use the inverse trigonometric function

C = arccos (\frac{- c ^{2} + a ^{2} + b ^{2}}{2 ab})

Calculate

C = arccos (\frac{- c ^{2} + a ^{2} + b ^{2}}{2 ab}) C = - arccos (\frac{- c ^{2} + a ^{2} + b ^{2}}{2 ab})

Add the period of 2 kπ, k \in Z to find all solutions

C = arccos (\frac{- c ^{2} + a ^{2} + b ^{2}}{2 ab}) + 2 kπ, k \in Z C = - arccos (\frac{- c ^{2} + a ^{2} + b ^{2}}{2 ab}) + 2 kπ, k \in Z

Solution

C = ⎩ ⎨ ⎧ arccos (\frac{- c ^{2} + a ^{2} + b ^{2}}{2 ab}) + 2 kπ - arccos (\frac{- c ^{2} + a ^{2} + b ^{2}}{2 ab}) + 2 kπ, k \in Z

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

Solve the equation

$Solve for C$

$Solve for a$

$Solve for b$