The GCF of 6, 42, and 12 is:
6.
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:
Area of circle that bounded by two raddii and the major arc is called major sector .....
Step-by-step explanation:
thank you
Answer:
C. 65º
Step-by-step explanation:
180º- (62+53)=65º
a=65º
180º-(65+62)=53º
b=53º
180º-(53+62)=65º
c=65º
