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3 years ago

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

Mathematics

2 answers:

Alex Ar [27]3 years ago

8 0

Answer:

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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3 years ago

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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:

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

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

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3 years ago