Consider the equation below.
1 answer:
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We have that
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer is
point (-5,12)
see the attached figure
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer is
point (-5,12)
see the attached figure
Answer:
- 150π ft²
- 10π ft.
Step-by-step explanation:
Area of the sector :
Finding the area given r = 30 ft. and θ = 60° :
⇒ Area = π × (30)² × 60/360
⇒ Area = π × 900/6
⇒ Area = 150π ft²
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Length of the arc :
Finding the arc length given r = 30 ft. and θ = 60° :
⇒ Arc Length = 2 × π × 30 × 60/360
⇒ Arc Length = 60/6 × π
⇒ Arc Length = 10π ft.