three trigonometric functions for a given angle are shown below csc theta = 13/12, sec theta = -13/5, cot theta = -5/12 what are
the coordinates of point (x,y) on The terminal ray of angle theta, assuming that the values above we’re not simplified.
2 answers:
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
You might be interested in
Answer:
(a)
Distance from player should be 13.82 feet or 36.2 feet
(b)
The ball will go over the net
Step-by-step explanation:
we are given
The ball follows a path given by the equation
where
x and y are measured in feet and the origin is on the court directly below where the player hits the ball
(a)
net height is 8 ft
so, we can set y=8
and then we can solve for x
we can use quadratic formula
So, distance from player should be 13.82 feet or 36.2 feet
(b)
we can plug x=30 and check whether y=8 ft
we know that
height of net is 8 ft
so, the ball will go over the net