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 ?
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i have an answer but im still kinda confused with the question
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.
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.
a, correct
b, slope is 1, or tan Theta=1
(tanTheta)=y/x