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Oksana_A [137]
4 years ago
11

Evaluate the function requested. Write your answer as a fraction in lowest terms. Triangle A B C. Angle C is 90 degrees. Hypoten

use A B is 35, side C B is 28, side C A is 21. Find sin A. a. Sine A = four-thirds c. sine A = three-fifths b. sine A = four-fifths d. sine A = five-fourths
Mathematics
1 answer:
Gekata [30.6K]4 years ago
3 0

Answer:

sine A = four-fifths

Step-by-step explanation:

You have a rectangle triangle ABC with hypotenuse AB 35, side CB 28 and side CA 21. The angle C is 90°.

In order to calculate sin A, you take into account that:

sinA=\frac{opposite\ side\ to\ A}{hypotenuse}

The opposite side to angle A is side CB = 28

Hypotenuse = 35

You replace the numeric values of the hypotenuse and side BC in the formula for sinA:

sinA=\frac{28}{35}

In order to simply the fraction, you divide both numerator and denominator by 7:

sinA=\frac{28/7}{35/7}=\frac{4}{5}

hence, the answer is

sine A = four-fifths

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Answer:

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Step-by-step explanation:

We are trying to evaluate this integral.

\int\limits_{0}^{4/\sqrt{2}}\,\,\int\limits_{y}^{\sqrt{16-y^2}}  xy \,\,dxdy

The first thing that we have to do is understand this region in the plane.

\{ (x,y) \in \mathbb{R} :  0\leq y \leq \frac{4}{\sqrt{2}} \,\, , y \leq x \leq \, \sqrt{16-y^2}   \}

If you graph it looks something like the photo I join.

Now we need to describe that same region in polar coordinates.

That same region in polar coordinates would be

\{ (r,\theta) : \,\, 0 \leq \theta \leq \frac{\pi}{4}  \,\,\, 0\leq r \leq 4  \}

Now remember that when we do the polar transformation we use the following formula

\int\limits_{a}^{b} \, \int\limits_{c}^{d}    f(x,y) \,dxdy =  \int\limits_{\theta_1}^{\theta_2} \, \int\limits_{r_1}^{r_2}    r* f(rcos(\theta),rsin(\theta))  \,drd\theta

Then our integral would be

\int\limits_{0}^{4/\sqrt{2}}\int\limits_{y}^{\sqrt{16-y^2}}  xy \, dxdy =  \int\limits_{0}^{\pi/4} \, \int\limits_{0}^{4}    r^3 cos(\theta)\sin(\theta) \,\,  drd\theta = 16

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