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An angle whose measure is 60⁰ is rotated more than halfway around a circle.
Since, we have to find the measure of angle.
As we already know that the angle of rotation about a circle is 360° therefore we have to find more than halfway of this angle.

Considering that an angle is rotated more than halfway around a circle be $\frac{1}{6}$

Multiplying $\frac{1}{6}$ with 360⁰
Therefore, it can show as $\frac{1}{6}$ ×360⁰
Which gives the result to be 60⁰
Hence, when an angle is measured 60⁰, it is rotating more than halfway around a circle.
A single rotation around a circle is equal to 360 degrees.
The measurement of an angle shows the magnitude and direction of the rotation of the angle from its initial position to the final position.
If the rotation is in a counterclockwise direction, it has an angle with positive measure. If the rotation is clockwise, it has an angle which gives negative measure.

4 0

3 years ago

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What is the sum of the measures of angle E, angle D, and angle ECD? A.76 B.104 C.128 D.180

Tcecarenko [31]

Answer:

serious??? D-180

Step-by-step explanation:

8 0

3 years ago

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 −

Korolek [52]

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv$

which has a value of 0, since each component integral is 0:

$\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0$

4 0

3 years ago