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2 years ago

7

Find cos theta if sin theta = 2/3. assume the terminal side of the angel falls in quadrant 2

1 answer:

kow [346]2 years ago

3 0

Well knowing that the terminal arm of the standard position angle is in quadrant 2, we can determine the reference angle, in quadrant 2, by simply taking the difference between 180 and whatever the angle is.

So ø reference = 180 - ø in standard position.

Regardless, the reference angle is in quadrant 2, we need to then label the sides of the reference triangle based on the opposite and hypotenuse.

Solve for adjacent side using Pythagoras theorem.

A^2 = C^2 - B^2
A^2 = 3^2 - 2^2
A^2 = 9 - 4
A^2 =5
A = sq root of 5.

Then write the cos ratio using the new side.

Cos ø =✔️5/3. Place a negative in front of cos ø as cos is negative in second quadrant.

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3 years ago

I have nooooo clue, please help

Aleksandr-060686 [28]

Answer:

case a) $x^{2}=3y$ ----> open up

case b) $x^{2}=-10y$ ----> open down

case c) $y^{2}=-2x$ ----> open left

case d) $y^{2}=6x$ ----> open right

Step-by-step explanation:

we know that

1) The general equation of a vertical parabola is equal to

$y=a(x-h)^{2}+k$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open upward and the vertex is a minimum

If a<0 ----> the parabola open downward and the vertex is a maximum

2) The general equation of a horizontal parabola is equal to

$x=a(y-k)^{2}+h$

where

a is a coefficient

(h,k) is the vertex

If a>0 ----> the parabola open to the right

If a<0 ----> the parabola open to the left

Verify each case

case a) we have

$x^{2}=3y$

so

$y=(1/3)x^{2}$

$a=(1/3)$

so

$a>0$

therefore

The parabola open up

case b) we have

$x^{2}=-10y$

so

$y=-(1/10)x^{2}$

$a=-(1/10)$

$a$

therefore

The parabola open down

case c) we have

$y^{2}=-2x$

so

$x=-(1/2)y^{2}$

$a=-(1/2)$

$a$

therefore

The parabola open to the left

case d) we have

$y^{2}=6x$

so

$x=(1/6)y^{2}$

$a=(1/6)$

$a>0$

therefore

The parabola open to the right

3 0

2 years ago