Answer:
[tex]\cos(\theta) = \dfrac{9}{41}[/tex]
Step-by-step explanation:
First, we can mark down the opposite and adjacent sides using our knowledge of the tangent ratio:
[tex]\tan(\theta)=\dfrac{9}{-40}=\dfrac{\text{opposite}}{\text{adjacent}}[/tex]
↓↓↓
- [tex]\text{opposite}=9[/tex]
- [tex]\text{adjacent} = -40}[/tex]
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Note that the given range for theta:
[tex]\dfrac{\pi}{2} \le \theta \le \pi[/tex]
tells us that the triangle is in the second quadrant. Thus, its adjacent (horizontal) side is the one that is negative, not its opposite (vertical) side.
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Next, we can solve for the triangle's hypotenuse with the Pythagorean Theorem:
[tex]a^2+b^2=c^2[/tex]
[tex]9^2 + (-40)^2 = c^2[/tex]
[tex]81 + 1600 = c^2[/tex]
[tex]c = \sqrt{1681}[/tex]
[tex]c = 14[/tex]
Finally, we can put this into the cosine ratio:
[tex]\cos(\theta)=\dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{9}{41}[/tex]
