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lord [1]
2 years ago
8

Angle A and angle B form a vertical angle pair. The measure of angle A is six less than twice times the measure of angle B. Find

the measure of each angle.
Mathematics
1 answer:
Studentka2010 [4]2 years ago
8 0

Angle A and Angle B are vertical angle pairs. The measure of angle A and B are equal because they are vertical pairs. Therefore, the measure of each angle is 6 degrees.

<h3>What is vertical angles?</h3>

Vertical angles are a pair of opposite angles formed by intersecting lines.

Vertical angles are congruent to each other . Vertical angles share the same vertex.

Therefore, angle A and angle B are congruent because they are vertical angle pair.

Therefore,

∠A≅∠B

let

∠B = x

Therefore,

∠A = 2x - 6

Therefore,

x = 2x - 6

x - 2x = - 6

-x = - 6

x = 6

The measure of each angle is 6 degrees.

learn more on vertical angles here: brainly.com/question/14753399

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Rufina [12.5K]

Answer:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

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We want to find the values between the interval (0, 2π) where the tangent line to the graph of y=sin(x)cos(x) is horizontal.

Since the tangent line is horizontal, this means that our derivative at those points are 0.

So, first, let's find the derivative of our function.

y=\sin(x)\cos(x)

Take the derivative of both sides with respect to x:

\frac{d}{dx}[y]=\frac{d}{dx}[\sin(x)\cos(x)]

We need to use the product rule:

(uv)'=u'v+uv'

So, differentiate:

y'=\frac{d}{dx}[\sin(x)]\cos(x)+\sin(x)\frac{d}{dx}[\cos(x)]

Evaluate:

y'=(\cos(x))(\cos(x))+\sin(x)(-\sin(x))

Simplify:

y'=\cos^2(x)-\sin^2(x)

Since our tangent line is horizontal, the slope is 0. So, substitute 0 for y':

0=\cos^2(x)-\sin^2(x)

Now, let's solve for x. First, we can use the difference of two squares to obtain:

0=(\cos(x)-\sin(x))(\cos(x)+\sin(x))

Zero Product Property:

0=\cos(x)-\sin(x)\text{ or } 0=\cos(x)+\sin(x)

Solve for each case.

Case 1:

0=\cos(x)-\sin(x)

Add sin(x) to both sides:

\cos(x)=\sin(x)

To solve this, we can use the unit circle.

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This only happens twice: at π/4 (45°) and at 5π/4 (225°).

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We have:

0=\cos(x)+\sin(x)

Subtract sine from both sides:

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Like the previous one, this also happens at the 45°. However, this times, it happens at 3π/4 and 7π/4.

At 3π/4, cosine is -√2/2, and sine is √2/2. If we divide by a negative, we will see that cos(x)=-sin(x).

At 7π/4, cosine is √2/2, and sine is -√2/2, thus making our equation true.

Therefore, our solution set is:

\{\frac{\pi}{4}, \frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\}

And we're done!

Edit: Small Mistake :)

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