not sure if im doing it right

[mathmom2015]

Find the exact value of cos θ if tan θ = -2/5
and θ is located in quadrant II

tan θ = -2/5 = opp/adj
-2²+-5²= square root 29^1/2
cos θ= -5 (29^1/2)/29

 

[serds]

seems to be ok.

 

[lookagain]

Find the exact value of cos θ if tan θ = -2/5
and θ is located in quadrant II

tan θ = -2/5 = opp/adj

-2²+-5²= square root 29^1/2 \(\displaystyle \ \ \ \) No, in the second quadrant for this problem, the opposite side is 2, and the adjacent side is -5, so the equation is: \(\displaystyle \ the \ \ square \ \ of \ \ the \ \ length \ \ of \ \ the \ \ hypotenuse \ = \ (2)^2 + (-5)^2 \ = \ 29.\)

Then the hypotenuse = \(\displaystyle \ \sqrt{29}. \)


cos θ = adj/hyp = \(\displaystyle \ \dfrac{-5}{\sqrt{29}} \ = \ \dfrac{-5\sqrt{29}}{29}.\)

.

 

[HallsofIvy]

Find the exact value of cos θ if tan θ = -2/5
and θ is located in quadrant II

tan θ = -2/5 = opp/adj
-2²+-5²= square root 29^1/2

This is written very awkwardly (-2)²+ (-5)²= 29. Then you take the square root of 29 to get the hypotenuse.
Strictly speaking -2²= -4. And you don't need both "square root" and the 1/2 power.

cos θ= -5 (29^1/2)/29

 

[Deleted member 4993]

Find the exact value of cos θ if tan θ = -2/5
and θ is located in quadrant II

tan θ = -2/5 = opp/adj
-2²+-5²= square root 29^1/2
cos θ= -5 (29^1/2)/29

Another way:

tan²(Θ) = sec²(Θ) - 1

sec²(Θ) = tan²(Θ) + 1 = 1 + 4/25 = 29/25

cos²(Θ) = 1/sec²(Θ) = 25/29

cos(Θ) = ± 5/√(29)

since Θ is in second quadrant cos(Θ) ≤ 0 → cos(Θ) = - 5/√(29) = - 5*√(29)/29

 

[lookagain]

This is written very awkwardly (-2)²+ (-5)²= 29. And that equation to the left is incorrect. See post #3. The opposite side is +2, while the adjacent side is -5, as theta is in Quadrant II.

So then \(\displaystyle tan(\theta) \) = opp/adj = (+2)/(-5) = -2/5.