Mathematics, 11.01.2020 18:31 owenbarrows

Suppose that vector v lies in quadrant ii, quadrant iii, or quadrant iv. how can you use trigonometric ratios to calculate the direction (i. e., angle) of the vector in each of these quadrants with respect to the positive x-axis? the angle between the vector and the positive x-axis will be greater than 90 in each case.

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Answer from: flowersthomas1969

I just had a test on vectors so I feel your pain! Explained in the picture above. Hope it helps C:
Edit: in picture I say tantheta = x/y sorry I meant y/x
Suppose that vector v lies in quadrant ii, quadrant iii, or quadrant iv. how can you use trigonometr

Suppose that vector v lies in quadrant ii, quadrant iii, or quadrant iv. how can you use trigonometr

Answer from: alicia186

The angles with respect the positive x-axis are given by the following formulas for each quadrant.

Second quadrant

$\theta =\tan^{-1} \left(\cfrac yx\right)+\pi\\\theta =\tan^{-1} \left(\cfrac yx\right)+180^\circ$

Third quadrant

$\theta =\tan^{-1} \left(\cfrac yx\right)+\pi\\\theta =\tan^{-1} \left(\cfrac yx\right)+180^\circ$

Fourth quadrant

$\theta =\tan^{-1} \left(\cfrac yx\right)+2\pi\\\theta =\tan^{-1} \left(\cfrac yx\right)+360^\circ$

Step-by-step explanation:

In order to calculate the angle in a general way, we always use the formula

$\theta =tan^{-1} \left( \cfrac yx \right)$

That will give us the angle of the associated reference triangle, the issue is that the the output of the inverse trigonometric function is in the range $[-\pi/2, \pi/2]$ and that does not give us the exact angle with respect the positive x-axis.

Thus we need to add some angle to get the correct angle with respect the positive x-axis.

Case second quadrant.

The output of $\theta =tan^{-1} \left( \cfrac yx \right)$ will be negative, since y is greater than 0 but x is less than zero.

To fix that, and get the angle on the second quadrant we can add either pi or 180 degrees, so we will get:

$\theta =\tan^{-1} \left(\cfrac yx\right)+\pi\\\theta =\tan^{-1} \left(\cfrac yx\right)+180^\circ$

Case third quadrant.

The output of $\theta =tan^{-1} \left( \cfrac yx \right)$ will be positive but it will be smaller than 90 degrees.

To fix that, and get the angle on the second quadrant we can add either pi or 180 degrees, so we will get:

$\theta =\tan^{-1} \left(\cfrac yx\right)+\pi\\\theta =\tan^{-1} \left(\cfrac yx\right)+180^\circ$

Case fourth quadrant

The output of $\theta =tan^{-1} \left( \cfrac yx \right)$ will be negative since y is less than 0 but x is positive.

To fix that we can get a positive coterminal angle, that means adding 2pi or 360 degrees, so we will get

$\theta =\tan^{-1} \left(\cfrac yx\right)+2\pi\\\theta =\tan^{-1} \left(\cfrac yx\right)+360^\circ$

Answer from: Quest

if you explain the the question more correctly i might be able to answer this better

step-by-step explanation:

Answer from: Quest

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

cretdit to source

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Suppose that vector v lies in quadrant ii, quadrant iii, or quadrant iv. how can you use trigonometr...