Ask question

Ask question

All categories

3 years ago

13

What is the value of tangent theta in the unit circle below? One-half StartFraction StartRoot 3 EndRoot Over 3 EndFraction Start

Fraction StartRoot 3 EndRoot Over 2 EndFraction StartRoot 3 EndRoot

2 answers:

horrorfan [7]3 years ago

6 0

Answer:

$\frac{1}{\sqrt{3} }$

Step-by-step explanation:

A unit circle is defined as a circle of unit radius (that is the radius of the circle is equal to 1). In trigonometry, the unit circle is a circle with a radius of 1, while centered at the origin (0, 0) in the Cartesian coordinate.

In the unit circle below, we can see that the line touches the unit circle at point $(\frac{\sqrt{3} }{2} ,\frac{1}{2})$ , therefore to find the tangent of theta, we use the formula:

$tan\theta = \frac{y}{x} \\\\tan\theta=\frac{\frac{1}{2} }{\frac{\sqrt{3} }{2} } =\frac{1}{2}*\frac{2}{\sqrt{3} } \\\\tan\theta=\frac{1}{\sqrt{3} }$

ZanzabumX [31]3 years ago

6 0

Answer:

sqrt3/3

Step-by-step explanation:

You might be interested in

Solve for the variable in 7⁄28 = 25⁄x A. 20 B. 100 C. 4 D. 5

LenaWriter [7]

7/28 = 25/x
x = (25 × 28)/7
x = (25 × 4)
x = 100

5 0

3 years ago

Read 2 more answers

Part A What is the electric field at the position (x1,y1)=(5.0 cm , 0 cm) in component form? Express your answer in terms of the

erik [133]

The complete Question is

A −12 nC charge is located at (x, y)=(1.0 cm, 0 cm).

Part A) What is the electric field at the position (x1, y1)=(5.0 cm, 0 cm) in component form?

Part B) What is the electric field at the position (x2, y2)=(−5.0 cm, 0 cm) in component form?

Part C) What is the electric field at the position (x3, y3)=(0 cm, 5.0 cm) in component form?

Express your answer in terms of the unit vectors i^ and j^. Use the 'unit vector' button to denote unit vectors in your answer.

Answer:

A) E = - 67500i + 0j N/C

B) E = 30000i + 0j N/C

C) E = 8143.18i - -40715.66j N/C

Step-by-step explanation:

Part A)

Magnitude of the charge = Q = -12 nC = $-12 \times 10^{-9}$ C

Since, its the negative charge, the direction of Electric Field lines will be directed toward the charge

Location of the charge = (1, 0)

We need to find the electric field at point (5, 0)

The formula for the magnitude of electric field due to a point charge is:

$E=\frac{kQ}{r^{2}}$

Here,

k = Coulomb's Law Constant = $9 \times 10^{9}$

Q = Magnitude of the charge

r = Distance between the charge and the point where we need to find the value of E

We can find r by using the Distance Formula.

So,

$r=\sqrt{(5-1)^{2}+(0-0)^{2}}=4$ cm = 0.04 m

Using these values in the formula, we get:

$E=\frac{9\times 10^{9} \times 12 \times 10^{-9}}{(0.04)^{2}}= 67500$ N/C

Since, two point (5, 0) is to the right of the given charge (as shown in the first image) i.e. in horizontal direction, all of the electric field experienced by it will be in horizontal direction and the vertical component would be zero. Also the direction of Electric field will be towards the charge i.e. in left direction so the x-component of Electric field will be negative.

Thus, we can write the value of E in vector form to be:

E = - 67500i + 0j N/C

Part B)

We need to find the Electric Field at point (-5, 0)

Using the similar procedure as used in the previous step, first we find r:

$r=\sqrt{(-5-1)^{2}+(0-0)^{2}}=6$ cm = 0.06

Using the values in the formula of Electric field, we get:

$E = \frac{9 \times 10^{9} \times 12 \times 10^{-9}}{(0.06)^2}=30000$ N/C

In this case again, the point is located in a horizontal direction to the given charge, so all the Electric Field experienced by it will be in horizontal direction and the vertical component will be zero. The direction of Electric field will be towards the charge i.e towards Right, so the x-component will be positive in this case.

So, value of the electric field in component form would be:

E = 30000i + 0j N/C

Part C)

We need to find the value of electric field at the point (0, 5). First we find the value of r:

$r=\sqrt{(1-0)^2+(0-5)^2}=\sqrt{26}=5.1$ cm = 0.051 m

Using the values in the formula of E, we get:

$E=\frac{kQ}{r^{2}}=\frac{9 \times 10^{9} \times 12 \times 10^{-9}}{(0.051)^{2}}=41522$ N/C

The point (0, 5) is neither exactly to the left or exactly up. So, for this point we need to find both the horizontal and vertical components as shown in the 2nd figure below.

From the triangle, we have the opposite and adjacent side to the angle, so using the tangent we can find the value of angle theta.

$tan(\theta)=\frac{5}{1}\\\theta=tan^{-1}(5)=78.69$

The two angles shown in the figure will be equal as there are alternate interior angles. Now the angle which E will make with positive x-axis will lie in the 4th quadrant as it lies below the horizontal line. So, the angle with positive x-axis would be:

360 - 78.69 = 281.31 degrees

Ex = E cos(θ) = 41522 cos(281.31) = 8143.18 N/C

Ey = E sin(θ) = 41522 sin(281.31) = -40715.66 N/C

So, in component form the Electric field will be:

E = 8143.18<em>i</em> - -40715.66<em>j</em>

3 0

3 years ago

Greg is designing a clock

klio [65]

What is the rest of the question or problem?

5 0

3 years ago

Help with this integral<br><br><img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%20%5Cfrac%7Bdx%7D%7Bx%5E%7B2%7D-4x-13%7D" id

charle [14.2K]

$x^2-4x-13=(x-2)^2-17$

$x-2=\sqrt{17}\sec y$
$\mathrm dx=\sqrt{17}\sec y\tan y\,\mathrm dy$

$\displaystyle\int\frac{\mathrm dx}{x^2-4x-13}=\int\frac{\sqrt{17}\sec y\tan y}{(\sqrt{17}\sec y)^2-17}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y\tan y}{\sec^2y-1}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y\tan y}{\tan^2y}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y}{\tan y}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\frac1{\cos y}}{\frac{\sin y}{\cos y}}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\csc y\,\mathrm dy$
$=-\dfrac1{\sqrt{17}}\ln|\csc y+\cot y|+C$

$\sec y=\dfrac{x-2}{\sqrt{17}}\iff y=\sec^{-1}\dfrac{x-2}{\sqrt{17}}$
$\implies\csc y=\dfrac{x-2}{\sqrt{(x-2)^2-17}}=\dfrac{x-2}{\sqrt{x^2-4x-13}}$
$\implies\cot y=\dfrac{\sqrt{17}}{\sqrt{(x-2)^2-17}}=\dfrac{\sqrt{17}}{\sqrt{x^2-4x-13}}$

$\displaystyle\int\frac{\mathrm dx}{x^2-4x-13}=-\dfrac1{\sqrt{17}}\ln\left|\frac{x-2+\sqrt{17}}{\sqrt{x^2-4x-13}}\right|+C$

6 0

4 years ago

PLZ HELP ASAP WILL GIVE BRAINLIEST

Paha777 [63]

The answer is 2,2 i already do the exam :D

6 0

4 years ago

Read 2 more answers