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ki77a [65]
2 years ago
15

Use the image below to answer the following question. Find the value of sin x° and cos y°. What relationship do the ratios of si

n x° and cos y° share?
A right triangle is shown with one leg measuring 12 and another leg measuring 5. The angle across from the leg measuring 5 is marked x degrees, and the angle across from the leg measuring 12 is marked y degrees.
Mathematics
2 answers:
juin [17]2 years ago
6 0

The computation of the angles show that the value of sin x and cos y will be 4/5 and 3/5 respectively.

How to calculate the angles?

From the information, we've to compute the length of the hypothenuse using Pythagoras theorem.

Therefore, 4² + 3² = hypothenuse²

hypothenuse² = 16 + 9

hypothenuse² = 25

hypothenuse = ✓25.

hypothenuse = 5

Therefore, sin x = opposite/hypothenuse = 4/5

cos y = adjacent/hypothenuse = 3/5

In conclusion, the values are 4/5 and 3/5.

Learn more about angles on:

Reika [66]2 years ago
5 0

The values of sin x° and cos y° are \frac{12}{13} and \frac{12}{13}. Since they are equal, the relationship between them is represented by sin x = cos y. This is obtained by using trigonometric ratios.

<h3>Pythagorean theorem:</h3>

This theorem states that the sum of squares of two legs of a right triangle is equal to the square of its hypotenuse. I.e.,

a^2+b^2=h^2

where a, and b are the two legs and 'h' is the hypotenuse of a right triangle.

<h3>Trigonometric ratios:</h3>

There are six trigonometric ratios. They are sine, cosine, tangent, secant, cosecant, and cotangent.

For a right-angle triangle,

These are ratios are calculated using the measure of an acute angle θ.

Such as

sin θ = \frac{opposite}{hypotenuse}

cos θ = \frac{adjacent}{hypotenuse}

tan θ = \frac{opposite}{adjacent} and the other three are evaluated from this.

<h3>Calculating the given ratios:</h3>

Given that:

A right triangle has one leg measuring 12 and another leg measuring 5.

The angle across from the leg measuring 5 is x° and the angle across from the leg measuring 12 is y°.

So, the triangle formed is shown in the figure below.

<h3>Step 1: Calculating the measurement value of hypotenuse:</h3>

The measurement of the two legs is 12 and 5. So, the hypotenuse is calculated as,

12² + 5² = h²

⇒ 144 + 25 = h²

⇒ 169 = h²

⇒ h = 13

Therefore, the hypotenuse measures 13.

<h3>Step2: Calculating the ratios sin x° and cos y°</h3>

From the trigonometric ratios, we know that,

sin x° = \frac{opposite}{hypotenuse} and

cos y°=\frac{adjacent}{hypotenuse}

For acute angle x°, the opposite side measures 12 and the hypotenuse is 13

So,

sin x° = \frac{12}{13}

Similarly, for acute angle y°, the adjacent side is 12 and the hypotenuse is 13

So,

cos y°=\frac{12}{13}

Thus, the two ratios are having the same value. Hence they are equal.

sin x = cos y.

Learn more about trigonometric ratios here:

brainly.com/question/24044139

#SPJ2

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Suppose Upper F Superscript prime Baseline left-parenthesis x right-parenthesis equals 3 x Superscript 2 Baseline plus 7 and Upp
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It looks like you're given

<em>F'(x)</em> = 3<em>x</em>² + 7

and

<em>F</em> (0) = 5

and you're asked to find <em>F(b)</em> for the values of <em>b</em> in the list {0, 0.1, 0.2, 0.5, 2.0}.

The first is done for you, <em>F</em> (0) = 5.

For the remaining <em>b</em>, you can solve for <em>F(x)</em> exactly by using the fundamental theorem of calculus:

F(x)=F(0)+\displaystyle\int_0^x F'(t)\,\mathrm dt

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F(x)=5+x^3+7x

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On the other hand, if you're expected to <em>approximate</em> <em>F</em> at the given <em>b</em>, you can use the linear approximation to <em>F(x)</em> around <em>x</em> = 0, which is

<em>F(x)</em> ≈ <em>L(x)</em> = <em>F</em> (0) + <em>F'</em> (0) (<em>x</em> - 0) = 5 + 7<em>x</em>

Then <em>F</em> (0) = 5, <em>F</em> (0.1) ≈ 5.7, <em>F</em> (0.2) ≈ 6.4, <em>F</em> (0.5) ≈ 8.5, and <em>F</em> (2.0) ≈ 19. Notice how the error gets larger the further away <em>b </em>gets from 0.

A <em>better</em> numerical method would be Euler's method. Given <em>F'(x)</em>, we iteratively use the linear approximation at successive points to get closer approximations to the actual values of <em>F(x)</em>.

Let <em>y(x)</em> = <em>F(x)</em>. Starting with <em>x</em>₀ = 0 and <em>y</em>₀ = <em>F(x</em>₀<em>)</em> = 5, we have

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<em>x</em>₂ = <em>x</em>₁ + 0.1 = 0.2

<em>y</em>₂ = <em>y</em>₁ + <em>F'(x</em>₁<em>)</em> (<em>x</em>₂ - <em>x</em>₁) = 5.7 + 7.03 (0.2 - 0.1)   →   <em>F</em> (0.2) ≈ 6.403

<em>x</em>₃ = <em>x</em>₂ + 0.3 = 0.5

<em>y</em>₃ = <em>y</em>₂ + <em>F'(x</em>₂<em>)</em> (<em>x</em>₃ - <em>x</em>₂) = 6.403 + 7.12 (0.5 - 0.2)   →   <em>F</em> (0.5) ≈ 8.539

<em>x</em>₄ = <em>x</em>₃ + 1.5 = 2.0

<em>y</em>₄ = <em>y</em>₃ + <em>F'(x</em>₃<em>)</em> (<em>x</em>₄ - <em>x</em>₃) = 8.539 + 7.75 (2.0 - 0.5)   →   <em>F</em> (2.0) ≈ 20.164

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