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liberstina [14]

3 years ago

8

A drill is used to dig down into the earth. After 2 hours, the drill is 12 meters below the surface. After 5 hours, the drill is

30 meters below the surface. Which equation represents the depth of the drill in meters, y, after x hours of drilling?

1 answer:

Zanzabum3 years ago

4 0

No idea to that answer sorry

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Given cos alpha = 8/17, alpha in quadrant IV, and sin beta = -24/25, beta in quadrant III, find sin(alpha-beta)

polet [3.4K]

Given $\cos\alpha=\frac{8}{17}$ , $\alpha$ is in Quadrant IV, $\sin\beta=-\frac{24}{25}$ , and $\beta$ is in Quadrant III, find $\sin(\alpha-\beta)$

We can use the angle subtraction formula of sine to answer this question.

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$

We already know that $\cos\alpha=\frac{8}{17}$ .

We can use the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ to find $\sin\alpha$ .

$\sin^2\alpha+(\frac{8}{17})^2=1 \\ \sin^2\alpha+\frac{64}{289}=1 \\ \sin^2\alpha=\frac{225}{289} \\ \\\sin\alpha=\pm\frac{15}{17}$

Since $\alpha$ is in Quadrant IV, and sine is represented as y value on the unit circle, we must assume the negative value $\sin\alpha=-\frac{15}{17}$ .

As similar process is then done with $\sin\beta=-\frac{24}{25}$ .

$(-\frac{24}{25})^2+\cos^2\beta=1 \\ \frac{576}{625}+\cos^2\beta=1 \\ \cos^2\beta=\frac{49}{625} \\ \\\cos\beta=\pm\frac{7}{25}$

And since $\beta$ is in Quadrant III, and cosine in represented as x value on the unit cercle, we must assume the negative value $\cos\beta=-\frac{7}{25}$ .

Now we can fill in our angle subtraction formula!

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta \\\\ \sin(\alpha-\beta)=(-\frac{15}{17}\times-\frac{7}{25})-(\frac{8}{17}\times-\frac{24}{25}) \\\\\sin(\alpha-\beta)=\frac{105}{425}-(-\frac{192}{425}) \\\\ \boxed{\sin(\alpha-\beta)=\frac{297}{425}}$

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