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Popular >

c^2=b^2+a^2-2(b)(a)cos(θ)

Solution

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

Solution

θ = arccos (- \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}) + 2 πn, θ = - arccos (- \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}) + 2 πn

Solution steps

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

Switch sides

b^{2} + a^{2} - 2 ba cos (θ) = c^{2}

Move

b^{2}

to the right side

b^{2} + a^{2} - 2 ba cos (θ) = c^{2}

Subtract

b^{2}

from both sides

b^{2} + a^{2} - 2 ba cos (θ) - b^{2} = c^{2} - b^{2}

Simplify

a^{2} - 2 ba cos (θ) = c^{2} - b^{2}

a^{2} - 2 ba cos (θ) = c^{2} - b^{2}

Move

a^{2}

to the right side

a^{2} - 2 ba cos (θ) = c^{2} - b^{2}

Subtract

a^{2}

from both sides

a^{2} - 2 ba cos (θ) - a^{2} = c^{2} - b^{2} - a^{2}

Simplify

- 2 ba cos (θ) = c^{2} - b^{2} - a^{2}

- 2 ba cos (θ) = c^{2} - b^{2} - a^{2}

Divide both sides by

- 2 ba; - 2 ba \neq = 0

- 2 ba cos (θ) = c^{2} - b^{2} - a^{2}

Divide both sides by

- 2 ba; - 2 ba \neq = 0

\frac{- 2 ba cos ( θ )}{- 2 ba} = \frac{c ^{2}}{- 2 ba} - \frac{b ^{2}}{- 2 ba} - \frac{a ^{2}}{- 2 ba}; - 2 ba \neq = 0

Simplify

\frac{- 2 ba cos ( θ )}{- 2 ba} = \frac{c ^{2}}{- 2 ba} - \frac{b ^{2}}{- 2 ba} - \frac{a ^{2}}{- 2 ba}

Simplify

\frac{- 2 ba cos ( θ )}{- 2 ba} : cos (θ)

\frac{- 2 ba cos ( θ )}{- 2 ba}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 ba cos ( θ )}{2 ba}

Divide the numbers:

\frac{2}{2} = 1

= \frac{ba cos ( θ )}{ba}

Cancel the common factor:

b

= \frac{a cos ( θ )}{a}

Cancel the common factor:

a

= cos (θ)

Simplify

\frac{c ^{2}}{- 2 ba} - \frac{b ^{2}}{- 2 ba} - \frac{a ^{2}}{- 2 ba} : - \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}

\frac{c ^{2}}{- 2 ba} - \frac{b ^{2}}{- 2 ba} - \frac{a ^{2}}{- 2 ba}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{c ^{2} - b ^{2} - a ^{2}}{- 2 ba}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}

cos (θ) = - \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}; - 2 ba \neq = 0

cos (θ) = - \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}; - 2 ba \neq = 0

cos (θ) = - \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}; - 2 ba \neq = 0

Apply trig inverse properties

cos (θ) = - \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}

General solutions for

cos (θ) = - \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

θ = arccos (- \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}) + 2 πn, θ = - arccos (- \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}) + 2 πn

θ = arccos (- \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}) + 2 πn, θ = - arccos (- \frac{c ^{2} - b ^{2} - a ^{2}}{2 ba}) + 2 πn

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