Can someone help me again with these math work?
2 answers:
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Answer:
Step-by-step explanation:
Given
Circle has radius
Area of the sector is given by
if is one-sixth of area of circle then
If is one-fourth of area of circle then
Answer:
The points are (3π/4, ) and (7π/4,
)
Step-by-step explanation:
Since the two equations r = 1 + cos(θ) and r = 1 − sin(θ) intersect, then
1 + cos(θ) = 1 − sin(θ)
collecting like terms
cos(θ) + sin(θ) = 1 - 1
cos(θ) + sin(θ) = 0
dividing through by cos(θ), we have
cos(θ)/cos(θ) + sin(θ)/cos(θ) = 0/cos(θ)
1 + tan(θ) = 0
tan(θ) = -1
Since tanθ is negative in both the second and fourth quadrant, we have
tan(π - θ) = 1 or tan(2π - θ) = 1
(π - θ) = tan⁻¹1 or (2π - θ) = tan⁻¹1
(π - θ) = π/4 or (2π - θ) = π/4
θ = π - π/4 = 3π/4 or θ = 2π - π/4 = 7π/4
Substituting these values into r = 1 + cos(θ), we have
r = 1 - cos(3π/4) or r = 1 - cos(7π/4)
r = 1 - (-1/√2) or r = 1 - 1/√2
r = 1 + √2/2 or r = 1 - √2/2
r = or r =
So, the points are (3π/4, ) and (7π/4,
)