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Question 1 of 3
Given that 0≥θ≥360, solve:
cosθ=-1√2
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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
The acute angle (θ’ is an angle less than 90° and is relative to the horizontal axis.
First, identify which quadrant theta may lie.
Since the given cosine value is negative, theta may be on Quadrant II or III.
Next, get the acute angle (theta’) for theta.
We can do this by matching the value of cos theta to an Exact Triangle.
\cos\theta=\frac{\color{#00880A}{\text{adjacent}}}{\color{#007DDC}{\text{hypotenuse}}}=-\frac{\color{#00880A}{1}}{\color{#007DDC}{\sqrt2}}
Since 1 is the adjacent side and sqrt2 is the hypotenuse, theta’=45°
Now, find the corresponding angle of the acute angle in both Quadrant II and III.
Therefore, theta=135°,225°
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Question 2 of 3
Given that 0≥theta≥360, solve:
sin theta=(sqrt3)/2
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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
The acute angle (theta’) is an angle less than 90° and is relative to the horizontal axis.
First, identify which quadrant theta may lie.
Since the given sine value is positive, theta may be on Quadrant I or II.
Next, get the acute angle (theta’) for theta.
We can do this by matching the value of sin theta to an Exact Triangle.
\sin\theta=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#007DDC}{\text{hypotenuse}}}=\frac{\color{#9a00c7}{\sqrt3}}{\color{#007DDC}{2}}
Since sqrt3 is the opposite side and 2 is the hypotenuse, theta’=60°
Now, find the corresponding angle of the acute angle in both Quadrant I and II.
Therefore, theta=60°,120°
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Question 3 of 3
Given that 0≥x≥360, solve:
tan x=-1
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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
The acute angle (theta’) is an angle less than 90° and is relative to the horizontal axis.
First, identify which quadrant x may lie.
Since the given tangent value is negative, x may be on Quadrant II or IV.
Next, get the acute angle for x.
We can do this by matching the value of tan x to an Exact Triangle.
\tan x=\frac{\color{#9a00c7}{\text{opposite}}}{\color{#00880A}{\text{adjacent}}}=-\frac{\color{#9a00c7}{1}}{\color{#00880A}{1}}
Since 1 is the opposite side and 1 is the adjacent, the acute angle is 45°
Now, find the corresponding angle of the acute angle in both Quadrant II and IV.
