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3 years ago

Evaluate each trigonometric equation for θ.

Mathematics

1 answer:

liberstina [14]3 years ago

4 0

Answer:

sinФ = 8/17

cosФ = 15/17

tan Ф = 8/15

csc Ф = 17/8

sec Ф = 17/15

cot Ф = 15/8

Step-by-step explanation:

Let us revise the trigonometry functions

Sin(x) = opposite/hypotenuse

Cos(x) = adjacent/hypoteouse

Tan(x) = opposite/adjacent

Csc(x) = hypotenuse/opposit

Sec(x) = hypotenues/adjacent

Cot(x) = adjacent/opposite

In the given figure

The opposite side to angle Ф = 8

The adjacent side to angle Ф = 15

Find the hypotenuse using Pythagoras' theorem

$Hypotenuse = \sqrt{8^{2}+15^{2} }$

$Hypotenuse = \sqrt{64+225}$

$Hypotenuse = \sqrt{289}$

$Hypotenuse = 17$

Let us use the rules above to find the trigonometry functions

sinФ = 8/17

cosФ = 15/17

tan Ф = 8/15

csc Ф = 17/8

sec Ф = 17/15

cot Ф = 15/8

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A parcel of land is in the shape of a rectangle with length L(in feet). sixty-five percent of the length is waterfront and 70 fe

OlgaM077 [116]

The length of the land = the length of the waterfront + the length of dry land

The length of the land = L
The length of the waterfront = 0.65L
The length of the dry land = 70 feet

L = 0.65L + 70
L - 0.65L = 70
0.35L = 70
L = 70 ÷ 0.35
L = 200 feet

8 0

3 years ago

Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .

Alik [6]

Answer:

-42

Step-by-step explanation:

The objective is to find the line integral of $F$ around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use <em>the Green's Theorem </em>to evaluate this integral. The rectangle is presented below.

We have that

$F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle$

Therefore,

$P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y$

Let's calculate the needed partial derivatives.

$P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0$

Thus,

$Q_x -P_y = 0 -2 = - 2$

Now, by the Green's theorem, we have

$\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x \Big|_{-3}^{4} = -42$