Trigonometry Examples

$(2, 5)$ , $(4, - 3)$

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.

$a⃗ \cdot b⃗ = 2 \cdot 4 + 5 \cdot - 3$

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 $2$ by $4$ .

$a⃗ \cdot b⃗ = 8 + 5 \cdot - 3$

Step 2.2.1.2

Multiply $5$ by $- 3$ .

$a⃗ \cdot b⃗ = 8 - 15$

Step 2.2.2

Subtract $15$ from $8$ .

$a⃗ \cdot b⃗ = - 7$

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.

$| a⃗ | = \sqrt{2^{2} + 5^{2}}$

Step 3.2

Simplify.

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Step 3.2.1

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

$| a⃗ | = \sqrt{4 + 5^{2}}$

Step 3.2.2

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

$| a⃗ | = \sqrt{4 + 25}$

Step 3.2.3

Add $4$ and $25$ .

$| a⃗ | = \sqrt{29}$

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.

$| b⃗ | = \sqrt{4^{2} + {(- 3)}^{2}}$

Step 4.2

Simplify.

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Step 4.2.1

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

$| b⃗ | = \sqrt{16 + {(- 3)}^{2}}$

Step 4.2.2

Raise $- 3$ to the power of $2$ .

$| b⃗ | = \sqrt{16 + 9}$

Step 4.2.3

Add $16$ and $9$ .

$| b⃗ | = \sqrt{25}$

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.

$θ = \arccos (\frac{- 7}{\sqrt{29} \cdot 5})$

Step 6

Simplify.

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Step 6.1

Move $5$ to the left of $\sqrt{29}$ .

$θ = \arccos (\frac{- 7}{5 \sqrt{29}})$

Step 6.2

Move the negative in front of the fraction.

$θ = \arccos (- \frac{7}{5 \sqrt{29}})$

Step 6.3

Multiply $\frac{7}{5 \sqrt{29}}$ by $\frac{\sqrt{29}}{\sqrt{29}}$ .

$θ = \arccos (- (\frac{7}{5 \sqrt{29}} \cdot \frac{\sqrt{29}}{\sqrt{29}}))$

Step 6.4

Combine and simplify the denominator.

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Step 6.4.1

Multiply $\frac{7}{5 \sqrt{29}}$ by $\frac{\sqrt{29}}{\sqrt{29}}$ .

$θ = \arccos (- \frac{7 \sqrt{29}}{5 \sqrt{29} \sqrt{29}})$

Step 6.4.2

Move $\sqrt{29}$ .

$θ = \arccos (- \frac{7 \sqrt{29}}{5 (\sqrt{29} \sqrt{29})})$

Step 6.4.3

Raise $\sqrt{29}$ to the power of $1$ .

$θ = \arccos (- \frac{7 \sqrt{29}}{5 ({\sqrt{29}}^{1} \sqrt{29})})$

Step 6.4.4

Raise $\sqrt{29}$ to the power of $1$ .

$θ = \arccos (- \frac{7 \sqrt{29}}{5 ({\sqrt{29}}^{1} {\sqrt{29}}^{1})})$

Step 6.4.5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$θ = \arccos (- \frac{7 \sqrt{29}}{5 {\sqrt{29}}^{1 + 1}})$

Step 6.4.6

Add $1$ and $1$ .

$θ = \arccos (- \frac{7 \sqrt{29}}{5 {\sqrt{29}}^{2}})$

Step 6.4.7

Rewrite ${\sqrt{29}}^{2}$ as $29$ .

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Step 6.4.7.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{29}$ as $29^{\frac{1}{2}}$ .

$θ = \arccos (- \frac{7 \sqrt{29}}{5 {(29^{\frac{1}{2}})}^{2}})$

Step 6.4.7.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$θ = \arccos (- \frac{7 \sqrt{29}}{5 \cdot 29^{\frac{1}{2} \cdot 2}})$

Step 6.4.7.3

Combine $\frac{1}{2}$ and $2$ .

$θ = \arccos (- \frac{7 \sqrt{29}}{5 \cdot 29^{\frac{2}{2}}})$

Step 6.4.7.4

Cancel the common factor of $2$ .

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Step 6.4.7.4.1

Cancel the common factor.

$θ = \arccos (- \frac{7 \sqrt{29}}{5 \cdot 29^{\frac{2}{2}}})$

Step 6.4.7.4.2

Rewrite the expression.

$θ = \arccos (- \frac{7 \sqrt{29}}{5 \cdot 29^{1}})$

Step 6.4.7.5

Evaluate the exponent.

$θ = \arccos (- \frac{7 \sqrt{29}}{5 \cdot 29})$

Step 6.5

Multiply $5$ by $29$ .

$θ = \arccos (- \frac{7 \sqrt{29}}{145})$

Step 6.6

Evaluate $\arccos (- \frac{7 \sqrt{29}}{145})$ .

$θ = 105.06848815$