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Question 1 of 4
Find the exact value of:
sin330°
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Positive Values in the Unit Circle
Quadrant I: All
Quadrant II: Sine only
Quadrant III: Tangent only
Quadrant IV: Cosine only
First, draw in the ray of 330°.
Next, get the acute angle (θ’) for 330°.
The acute angle is an angle less than 90° and is relative to the horizontal axis.
Based on this diagram, simply subtract 330° from 360° to get the acute angle.
Now, identify if sine is positive or negative in the quadrant where the ray lies, Quadrant IV.
The sine value is negative in Quadrant IV.
Finally, we can get the exact value of sin330° by solving for -sin30° using Exact Triangle Ratios.
sin330° |
= |
-sin30° |
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= |
−oppositehypotenuse |
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= |
−12 |
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Question 2 of 4
Find the exact value of:
tan120°
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Positive Values in the Unit Circle
Quadrant I: All
Quadrant II: Sine only
Quadrant III: Tangent only
Quadrant IV: Cosine only
First, draw in the ray of 120°.
Next, get the acute angle (θ’) for 120°.
The acute angle is an angle less than 90° and is relative to the horizontal axis.
Based on this diagram, simply subtract 120° from 180° to get the acute angle.
Now, identify if tangent is positive or negative in the quadrant where the ray lies, Quadrant II.
The tangent value is negative in Quadrant II.
Finally, we can get the exact value of tan120° by solving for -tan60° using Exact Triangle Ratios.
tan120° |
= |
-tan60° |
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= |
−oppositeadjacent |
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= |
−√31 |
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= |
-√3 |
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Question 3 of 4
Find the exact value of:
sec315°
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Positive Values in the Unit Circle
Quadrant I: All
Quadrant II: Sine only
Quadrant III: Tangent only
Quadrant IV: Cosine only
First, draw in the ray of 315°.
Next, get the acute angle (θ’) for 315°.
The acute angle is an angle less than 90° and is relative to the horizontal axis.
Based on this diagram, simply subtract 315° from 360° to get the acute angle.
Finally, we can get the exact value of sec315° by solving for sec45° using Exact Triangle Ratios.
Note that since secθ=1cosθ and cosine is positive in Quadrant IV, secant is also positive.
sec315° |
= |
sec45° |
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= |
1cos45° |
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= |
1oppositeadjacent |
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= |
11√2 |
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= |
√2 |
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Question 4 of 4
Find the exact value of:
cot150°
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Positive Values in the Unit Circle
Quadrant I: All
Quadrant II: Sine only
Quadrant III: Tangent only
Quadrant IV: Cosine only
First, draw in the ray of 150°.
Next, get the acute angle (θ’) for 150°.
The acute angle is an angle less than 90° and is relative to the horizontal axis.
Based on this diagram, simply subtract 150° from 180° to get the acute angle.
Now, identify if cotangent is positive or negative in the quadrant where the ray lies, Quadrant II.
The cotangent value is negative in Quadrant II.
Finally, we can get the exact value of cot150° by solving for -cot30° using Exact Triangle Ratios.
Note that cotθ=1tanθ
cot150° |
= |
-cot30° |
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= |
-1tan30° |
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= |
−1oppositeadjacent |
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= |
−11√3 |
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= |
-√3 |