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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
A regular trapezoid is shown in the picture attached.
We know that:
DC = minor base = 4
AB = major base = 7
AD = BC = lateral sides or legs = 5
Since the two legs have the same length, the trapezoid is isosceles and we can calculate AH by the formula:
AH = (AB - DC) ÷ 2
= (7 - 5) ÷ 2
= 2 ÷ 2
= 1
Now, we can apply the Pythagorean theorem in order to calculate DH:
DH = √(AD² - AH²)
= √(5² - 1²)
= √(25 - 1)
= √24
= 2√6
Last, we have all the information needed in order to calculate the area by the formula:
A = (7 + 5) × 2√6 ÷ 2
= 12√6
The area of the regular trapezoid is 12√6 square units.
We know that:
DC = minor base = 4
AB = major base = 7
AD = BC = lateral sides or legs = 5
Since the two legs have the same length, the trapezoid is isosceles and we can calculate AH by the formula:
AH = (AB - DC) ÷ 2
= (7 - 5) ÷ 2
= 2 ÷ 2
= 1
Now, we can apply the Pythagorean theorem in order to calculate DH:
DH = √(AD² - AH²)
= √(5² - 1²)
= √(25 - 1)
= √24
= 2√6
Last, we have all the information needed in order to calculate the area by the formula:
A = (7 + 5) × 2√6 ÷ 2
= 12√6
The area of the regular trapezoid is 12√6 square units.