One is 3pi/2 - 2pi = -pi/2
also -7pi/2
Let ABC be a triangle in the 3rd quadrant, right-angled at B.
So, AB-> Perpendicular BC -> Base AC -> Hypotenuse.
Given: sinθ=-3/5 cosecθ=-5/3
According to Pythagorean theorem, square of the hypotenuse is equal to the sum of square of the other two sides.
Therefore in triangle ABC, 〖AC〗^2=〖AB〗^2+〖BC〗^2 ------
--(1)
Since sinθ=Perpendicular/Hypotenuse ,
AC=5 and AB=3
Substituting these values in equation (1)
〖BC〗^2=〖AC〗^2-〖AB〗^2
〖BC〗^2=5^2-3^2
〖BC〗^2=25-9
〖BC〗^2=16
BC=4 units
Since the triangle is in the 3rd quadrant, all trigonometric ratios, except tan
and cot are negative.
So,cosθ=Base/Hypotenuse Cosθ=-4/5
secθ=Hypotnuse/Base secθ=-5/4
tanθ=Perpendicular/Base tanθ=3/4
cotθ=Base/Perpendicular cotθ=4/3
13x=8x+25
5x=25
X=5
5 is the answer
Answer:
C) 3,-2,-7,-12
Step-by-step explanation:
A sequence is defined recursively using the formula f(n+1)=f(n)-5. Therefore, 3,-2,-7,-12 is the sequence that could be generated using this formula.
