Trigonometry Examples

$(0, 4)$ , $(0, - 5)$

Step 1

Use the dot product formula to find the angle between two vectors.

$θ = \arccos (\frac{a⃗ \cdot b⃗}{| a⃗ | | b⃗ |})$

Step 2

Find the dot product.

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Step 2.1

The dot product of two vectors is the sum of the products of the their components.

$a⃗ \cdot b⃗ = 0 \cdot 0 + 4 \cdot - 5$

Step 2.2

Simplify.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Multiply $0$ by $0$ .

$a⃗ \cdot b⃗ = 0 + 4 \cdot - 5$

Step 2.2.1.2

Multiply $4$ by $- 5$ .

$a⃗ \cdot b⃗ = 0 - 20$

Step 2.2.2

Subtract $20$ from $0$ .

$a⃗ \cdot b⃗ = - 20$

Step 3

Find the magnitude of $a⃗$ .

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Step 3.1

The norm is the square root of the sum of squares of each element in the vector.

$| a⃗ | = \sqrt{0^{2} + 4^{2}}$

Step 3.2

Simplify.

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Step 3.2.1

Raising $0$ to any positive power yields $0$ .

$| a⃗ | = \sqrt{0 + 4^{2}}$

Step 3.2.2

Raise $4$ to the power of $2$ .

$| a⃗ | = \sqrt{0 + 16}$

Step 3.2.3

Add $0$ and $16$ .

$| a⃗ | = \sqrt{16}$

Step 3.2.4

Rewrite $16$ as $4^{2}$ .

$| a⃗ | = \sqrt{4^{2}}$

Step 3.2.5

Pull terms out from under the radical, assuming positive real numbers.

$| a⃗ | = 4$

Step 4

Find the magnitude of $b⃗$ .

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Step 4.1

The norm is the square root of the sum of squares of each element in the vector.

$| b⃗ | = \sqrt{0^{2} + {(- 5)}^{2}}$

Step 4.2

Simplify.

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Step 4.2.1

Raising $0$ to any positive power yields $0$ .

$| b⃗ | = \sqrt{0 + {(- 5)}^{2}}$

Step 4.2.2

Raise $- 5$ to the power of $2$ .

$| b⃗ | = \sqrt{0 + 25}$

Step 4.2.3

Add $0$ and $25$ .

$| b⃗ | = \sqrt{25}$

Step 4.2.4

Rewrite $25$ as $5^{2}$ .

$| b⃗ | = \sqrt{5^{2}}$

Step 4.2.5

Pull terms out from under the radical, assuming positive real numbers.

$| b⃗ | = 5$

Step 5

Substitute the values into the formula.

$θ = \arccos (\frac{- 20}{4 \cdot 5})$

Step 6

Simplify.

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Step 6.1

Cancel the common factor of $- 20$ and $4$ .

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Step 6.1.1

Factor $4$ out of $- 20$ .

$θ = \arccos (\frac{4 \cdot - 5}{4 \cdot 5})$

Step 6.1.2

Cancel the common factors.

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Step 6.1.2.1

Factor $4$ out of $4 \cdot 5$ .

$θ = \arccos (\frac{4 \cdot - 5}{4 (5)})$

Step 6.1.2.2

Cancel the common factor.

$θ = \arccos (\frac{4 \cdot - 5}{4 \cdot 5})$

Step 6.1.2.3

Rewrite the expression.

$θ = \arccos (\frac{- 5}{5})$

Step 6.2

Cancel the common factor of $- 5$ and $5$ .

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Step 6.2.1

Factor $5$ out of $- 5$ .

$θ = \arccos (\frac{5 \cdot - 1}{5})$

Step 6.2.2

Cancel the common factors.

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Step 6.2.2.1

Factor $5$ out of $5$ .

$θ = \arccos (\frac{5 \cdot - 1}{5 (1)})$

Step 6.2.2.2

Cancel the common factor.

$θ = \arccos (\frac{5 \cdot - 1}{5 \cdot 1})$

Step 6.2.2.3

Rewrite the expression.

$θ = \arccos (\frac{- 1}{1})$

Step 6.2.2.4

Divide $- 1$ by $1$ .

$θ = \arccos (- 1)$

Step 6.3

The exact value of $\arccos (- 1)$ is $180$ .

$θ = 180$