Solved on Feb 11, 2024

Find $\tan (\theta)$ for a triangle with $h=60$ and $\ell=12$ , where $\ell$ is the opposite and $h$ is the hypotenuse.

STEP 1

Assumptions
1. In the given right triangle, $\ell=12$ represents the length of the side opposite to angle $\theta$ .
2. The hypotenuse of the triangle is given by $h=60$ .
3. We are asked to find $\tan(\theta)$ , which is the ratio of the opposite side to the adjacent side in a right triangle.
4. Since we have the hypotenuse and the opposite side, we will need to find the length of the adjacent side to calculate $\tan(\theta)$ .

STEP 2

We can use the Pythagorean theorem to find the length of the adjacent side, which we will call $a$ . The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
$h^2 = \ell^2 + a^2$

STEP 3

Now, substitute the given values for $h$ and $\ell$ into the Pythagorean theorem to find $a^2$ .
$60^2 = 12^2 + a^2$

STEP 4

Calculate the squares of the given numbers.
$3600 = 144 + a^2$

STEP 5

Subtract $144$ from both sides of the equation to solve for $a^2$ .
$a^2 = 3600 - 144$

STEP 6

Calculate the value of $a^2$ .
$a^2 = 3456$

STEP 7

Take the square root of both sides to find the length of the adjacent side $a$ .
$a = \sqrt{3456}$

STEP 8

Simplify the square root by finding the prime factors of $3456$ or recognizing perfect squares within $3456$ .
$a = \sqrt{64 \times 54}$

STEP 9

Further simplify the square root by separating the perfect square.
$a = \sqrt{64} \times \sqrt{54}$

STEP 10

Calculate the square root of the perfect square and simplify the square root of $54$ .
$a = 8 \times \sqrt{54}$

STEP 11

Recognize that $54$ can be further factored into $9 \times 6$ to simplify the square root.
$a = 8 \times \sqrt{9 \times 6}$

STEP 12

Separate the square root of the perfect square.
$a = 8 \times \sqrt{9} \times \sqrt{6}$

STEP 13

Calculate the square root of the perfect square.
$a = 8 \times 3 \times \sqrt{6}$

STEP 14

Multiply the constants outside the square root.
$a = 24 \times \sqrt{6}$

STEP 15

Now that we have the length of the adjacent side $a$ , we can calculate $\tan(\theta)$ .
$\tan(\theta) = \frac{\ell}{a}$

STEP 16

Substitute the values of $\ell$ and $a$ into the formula for $\tan(\theta)$ .
$\tan(\theta) = \frac{12}{24 \times \sqrt{6}}$

STEP 17

Simplify the fraction by dividing both the numerator and the denominator by $12$ .
$\tan(\theta) = \frac{1}{2 \times \sqrt{6}}$

STEP 18

Rationalize the denominator by multiplying the numerator and the denominator by $\sqrt{6}$ .
$\tan(\theta) = \frac{1 \times \sqrt{6}}{2 \times \sqrt{6} \times \sqrt{6}}$

STEP 19

Simplify the denominator.
$\tan(\theta) = \frac{\sqrt{6}}{2 \times 6}$

STEP 20

Calculate the denominator.
$\tan(\theta) = \frac{\sqrt{6}}{12}$
The exact value of $\tan(\theta)$ in the given triangle is $\frac{\sqrt{6}}{12}$ .

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Find tan⁡(θ)\tan (\theta)tan(θ) for a triangle with h=60h=60h=60 and ℓ=12\ell=12ℓ=12, where ℓ\ellℓ is the opposite and hhh is the hypotenuse.

STEP 1

STEP 2

We can use the Pythagorean theorem to find the length of the adjacent side, which we will call aaa. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.h2=ℓ2+a2h^2 = \ell^2 + a^2h2=ℓ2+a2

STEP 3

Now, substitute the given values for hhh and ℓ\ellℓ into the Pythagorean theorem to find a2a^2a2.602=122+a260^2 = 12^2 + a^2602=122+a2

STEP 4

Calculate the squares of the given numbers.3600=144+a23600 = 144 + a^23600=144+a2

STEP 5

Subtract 144144144 from both sides of the equation to solve for a2a^2a2.a2=3600−144a^2 = 3600 - 144a2=3600−144

STEP 6

Calculate the value of a2a^2a2.a2=3456a^2 = 3456a2=3456

STEP 7

Take the square root of both sides to find the length of the adjacent side aaa.a=3456a = \sqrt{3456}a=3456​

STEP 8

Simplify the square root by finding the prime factors of 345634563456 or recognizing perfect squares within 345634563456.a=64×54a = \sqrt{64 \times 54}a=64×54​

STEP 9

Further simplify the square root by separating the perfect square.a=64×54a = \sqrt{64} \times \sqrt{54}a=64​×54​

STEP 10

Calculate the square root of the perfect square and simplify the square root of 545454.a=8×54a = 8 \times \sqrt{54}a=8×54​

STEP 11

Recognize that 545454 can be further factored into 9×69 \times 69×6 to simplify the square root.a=8×9×6a = 8 \times \sqrt{9 \times 6}a=8×9×6​

STEP 12

Separate the square root of the perfect square.a=8×9×6a = 8 \times \sqrt{9} \times \sqrt{6}a=8×9​×6​

STEP 13

Calculate the square root of the perfect square.a=8×3×6a = 8 \times 3 \times \sqrt{6}a=8×3×6​

STEP 14

Multiply the constants outside the square root.a=24×6a = 24 \times \sqrt{6}a=24×6​

STEP 15

Now that we have the length of the adjacent side aaa, we can calculate tan⁡(θ)\tan(\theta)tan(θ).tan⁡(θ)=ℓa\tan(\theta) = \frac{\ell}{a}tan(θ)=aℓ​

STEP 16

Substitute the values of ℓ\ellℓ and aaa into the formula for tan⁡(θ)\tan(\theta)tan(θ).tan⁡(θ)=1224×6\tan(\theta) = \frac{12}{24 \times \sqrt{6}}tan(θ)=24×6​12​

STEP 17

Simplify the fraction by dividing both the numerator and the denominator by 121212.tan⁡(θ)=12×6\tan(\theta) = \frac{1}{2 \times \sqrt{6}}tan(θ)=2×6​1​

STEP 18

Rationalize the denominator by multiplying the numerator and the denominator by 6\sqrt{6}6​.tan⁡(θ)=1×62×6×6\tan(\theta) = \frac{1 \times \sqrt{6}}{2 \times \sqrt{6} \times \sqrt{6}}tan(θ)=2×6​×6​1×6​​

STEP 19

Simplify the denominator.tan⁡(θ)=62×6\tan(\theta) = \frac{\sqrt{6}}{2 \times 6}tan(θ)=2×66​​

STEP 20

Calculate the denominator.tan⁡(θ)=612\tan(\theta) = \frac{\sqrt{6}}{12}tan(θ)=126​​The exact value of tan⁡(θ)\tan(\theta)tan(θ) in the given triangle is 612\frac{\sqrt{6}}{12}126​​.

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Find $\tan (\theta)$ for a triangle with $h=60$ and $\ell=12$ , where $\ell$ is the opposite and $h$ is the hypotenuse.

We can use the Pythagorean theorem to find the length of the adjacent side, which we will call $a$ . The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
$h^2 = \ell^2 + a^2$

Now, substitute the given values for $h$ and $\ell$ into the Pythagorean theorem to find $a^2$ .
$60^2 = 12^2 + a^2$

Calculate the squares of the given numbers.
$3600 = 144 + a^2$

Subtract $144$ from both sides of the equation to solve for $a^2$ .
$a^2 = 3600 - 144$

Calculate the value of $a^2$ .
$a^2 = 3456$

Take the square root of both sides to find the length of the adjacent side $a$ .
$a = \sqrt{3456}$

Simplify the square root by finding the prime factors of $3456$ or recognizing perfect squares within $3456$ .
$a = \sqrt{64 \times 54}$

Further simplify the square root by separating the perfect square.
$a = \sqrt{64} \times \sqrt{54}$

Calculate the square root of the perfect square and simplify the square root of $54$ .
$a = 8 \times \sqrt{54}$

Recognize that $54$ can be further factored into $9 \times 6$ to simplify the square root.
$a = 8 \times \sqrt{9 \times 6}$

Separate the square root of the perfect square.
$a = 8 \times \sqrt{9} \times \sqrt{6}$

Calculate the square root of the perfect square.
$a = 8 \times 3 \times \sqrt{6}$

Multiply the constants outside the square root.
$a = 24 \times \sqrt{6}$

Now that we have the length of the adjacent side $a$ , we can calculate $\tan(\theta)$ .
$\tan(\theta) = \frac{\ell}{a}$

Substitute the values of $\ell$ and $a$ into the formula for $\tan(\theta)$ .
$\tan(\theta) = \frac{12}{24 \times \sqrt{6}}$

Simplify the fraction by dividing both the numerator and the denominator by $12$ .
$\tan(\theta) = \frac{1}{2 \times \sqrt{6}}$

Rationalize the denominator by multiplying the numerator and the denominator by $\sqrt{6}$ .
$\tan(\theta) = \frac{1 \times \sqrt{6}}{2 \times \sqrt{6} \times \sqrt{6}}$

Simplify the denominator.
$\tan(\theta) = \frac{\sqrt{6}}{2 \times 6}$

Calculate the denominator.
$\tan(\theta) = \frac{\sqrt{6}}{12}$
The exact value of $\tan(\theta)$ in the given triangle is $\frac{\sqrt{6}}{12}$ .