g(θ) = 20θ − 5 tan θ
To find out critical points we take first derivative and set it =0
g(θ) = 20θ − 5 tan θ
g'(θ) = 20 − 5 sec^2(θ)
Now we set derivative =0
20 − 5 sec^2(θ)=0
Subtract 20 from both sides
− 5 sec^2(θ)=0 -20
Divide both sides by 5
sec^2(θ)= 4
Take square root on both sides
sec(θ)= -2 and sec(θ)= +2
sec can be written as 1/cos
so sec(θ)= -2 can be written as cos(θ)= -1/2
Using unit circle the value of θ is 
sec(θ)= 2 can be written as cos(θ)=1/2
Using unit circle the value of θ is 
For general solution we add 2npi
So critical points are

Answer:
No it does not equal a right triangle.
Answer:
k= -18
Step-by-step explanation:
5 = (k/6) + 8 so 5-8 = k/6
-3=k/6 then -3 multiplied by 6 = k which is -18 = k
-18/6= -3
-3+8=5
cot x = cos x / sin x
cot x = .85 / .52 = 1.6346