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JulijaS [17]
3 years ago
11

Help!!! Solve for x!! Pls!!!

Mathematics
2 answers:
Alex Ar [27]3 years ago
8 0

Answer:

57

Step-by-step explanation:

The bottom line of the triangle is 180 degrees. Therefore, 2y + 4y - 36 has to equal 180. So, once you solve this linear equation, you will find that y equals 36. A triangle's angles always equal a total of 180 degrees. So, x + 51 + 2y has to equal 180, as well. Since we know y, substitute in 36 for 2y. Now the equation is 51 + 72 + x = 180. Solve for x. X is 57.

Rudiy273 years ago
7 0

Answer:

57°

Step-by-step explanation:

The angles that are at the bottom middle are on a straight line, so they must add up to 180°. In order to determine y, use the equation 180 = 2y + 4y -36. After solving for y, you should be left with 36 = y. Now, we can fill in the bottom right angle in the left triangle. This would be 36 times 2, or 72. In triangles, all angles always add up to 180°, so, we can use the equation 180 = x + 51 + 72. Once solved for x, this will leave you with 57 = x, our final answer.

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Mamont248 [21]

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Step-by-step explanation:

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What are the perimeter and the area of the square? <br>80<br>200<br>400<br>8√5<br>16√5<br>32√5​
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3 years ago
Please help me for the love of God if i fail I have to repeat the class
Elena-2011 [213]

\theta is in quadrant I, so \cos\theta>0.

x is in quadrant II, so \sin x>0.

Recall that for any angle \alpha,

\sin^2\alpha+\cos^2\alpha=1

Then with the conditions determined above, we get

\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35

and

\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}

Now recall the compound angle formulas:

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta

\sin2\alpha=2\sin\alpha\cos\alpha

\cos2\alpha=\cos^2\alpha-\sin^2\alpha

as well as the definition of tangent:

\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}

Then

1. \sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}

2. \cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}

3. \tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}

4. \sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}

5. \cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}

6. \tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7

7. A bit more work required here. Recall the half-angle identities:

\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2

\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2

\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}

Because x is in quadrant II, we know that \dfrac x2 is in quadrant I. Specifically, we know \dfrac\pi2, so \dfrac\pi4. In this quadrant, we have \tan\dfrac x2>0, so

\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32

8. \sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}

6 0
3 years ago
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