The line with equation a + 2b = 0 coincides with the terminal side of an angle θ in standard position and cos θ<0 .

Question

The line with equation a + 2b = 0 coincides with the terminal side of an angle θ in standard position and cos θ<0 .

What is the value of sinθ ?

−1/2 <my answer

5√

5√/5

−25√/5

The line with equation x−y=0 coincides with the terminal side of an angle θ in standard position in Quadrant III .

What is the value of tanθ ?

Enter your answer in the box.

i have an answer but im still kinda confused with the question

Answer 1

For the first question, we know that the line with equation a + 2b = 0 coincides with the terminal side of an angle θ in standard position and cos θ < 0.

To find the value of sinθ, we can use the Pythagorean identity:
sin^2θ + cos^2θ = 1

Since cos θ < 0, we know that sinθ is positive (since it cannot be zero) and θ lies in either Quadrant II or Quadrant III. In these quadrants, sinθ is positive.

Therefore, the value of sinθ is 1/2.

For the second question, we know that the line with equation x - y = 0 coincides with the terminal side of an angle θ in standard position in Quadrant III.

To find the value of tanθ, we can use the tangent trigonometric ratio:
tanθ = sinθ / cosθ

Since the line lies in Quadrant III, both x and y coordinates are negative. Therefore, tanθ is positive in this quadrant.

So, the value of tanθ is 1.

Answer 2

To find the value of sinθ, we need to use the equation a + 2b = 0 and the given information that cosθ < 0.

To start, let's rearrange the equation a + 2b = 0 to solve for b:

a + 2b = 0
2b = -a
b = -a/2

Now, we know that cosθ = a and sinθ = b. Since cosθ < 0, we can conclude that a < 0.

Next, we substitute b = -a/2 into sinθ = b:

sinθ = -a/2

Since a < 0, we divide both sides of the equation by -1 to preserve the inequality sign:

-sinθ = a/2

Therefore, the value of sinθ is -a/2.

For the second question,

The line with equation x - y = 0 coincides with the terminal side of an angle θ in standard position in Quadrant III.

To find the value of tanθ, we can rearrange the equation x - y = 0 to solve for y:

x - y = 0
y = x

Now, we know that tanθ = y/x. Since we are in Quadrant III, both x and y will be negative. Therefore, we can write:

tanθ = (-y)/(-x) = y/x = 1

The value of tanθ in this case is 1.

Answer 3

a, correct

b, slope is 1, or tan Theta=1
(tanTheta)=y/x