Begin with cos(θ) = 5/13, θ in Quadrant IV
you should distinguish the 5-12-13 right-angled triangle
and then cosØ = adjacent/hypotenuse
x = 5, r = 13 , y = -12, since Ø is in IV
and sinØ = -12/13
also tan(ϕ) = −√15 = -√15/1 = y/x and ϕ is in II,
y = √15 , x = -1
r^2 = x^2 + y^2 = 15+1 = 16
r = 4
sinϕ = √15/4 , cosϕ = -1/4
you must know that:
cos(θ − ϕ) = cosθcosϕ + sinθsinϕ
= (5/13)(-1/4) + (-12/13)(√15/4)
= -5/52 - 12√15/52
= (-5-12√15)/52
Answer:
Step-by-step explanation:
1 Equation is given, so solving for "x" would mean an expression for "x", where
x = SOMETHING
First, we need to isolate x, so we need to take "8y" to the other side, we have:
Now, since we don't have "x" isolated yet, we need to take out the 2, so we will divide both sides by 2 to get x isolated. Shown below:
So, x has this expression
Yes it's true because:
TSR and RQP is alternating tringle
TSR & PQR is perpendicular
Answer:
50 degrees
Step-by-step explanation:
vertical angles are the same
angle LPM = 180-60-70
angle LPM = 180-130 = 50
angle LPM = angle KPN
50=angle KPN
Answer: (ln10)/6=t
Step-by-step explanation: e is an irrational number, like pi. Therefore the only variable to solve for is "t." Divide both sides by 4, then subtract (1/4) from both sides to get 10. Then take the natural log of both sides because you get rid of e on the left side, and you have the variable by itself. Natural log, or ln, represents loge base e, so dividing both sides by 6, we get t! I hope this helps!
