Gradient of line,m=(17-6)/(20-0)=11/20
finding equation of line using a point (20,17)
y-17=11/20(x-20)
y-17=(11/20x)-11
y=11/20x+(17-11)
y=(11/20)*x+6
Answer: (a) x + 2y = -1 and 3x + y = 1
<u>Step-by-step explanation:</u>
I am not sure what the purpose was for the colored lines but I included them on the graph (below).
The answer would be n = 1
By definition of tangent,
tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)
Recall the double angle identities:
sin(2<em>θ</em>) = 2 sin(<em>θ</em>) cos(<em>θ</em>)
cos(2<em>θ</em>) = cos²(<em>θ</em>) - sin²(<em>θ</em>) = 2 cos²(<em>θ</em>) - 1
where the latter equality follows from the Pythagorean identity, cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1. From this identity we can solve for the unknown value of sin(<em>θ</em>):
sin(<em>θ</em>) = ± √(1 - cos²(<em>θ</em>))
and the sign of sin(<em>θ</em>) is determined by the quadrant in which the angle terminates.
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We're given that <em>θ</em> belongs to the third quadrant, for which both sin(<em>θ</em>) and cos(<em>θ</em>) are negative. So if cos(<em>θ</em>) = -4/5, we get
sin(<em>θ</em>) = - √(1 - (-4/5)²) = -3/5
Then
tan(2<em>θ</em>) = sin(2<em>θ</em>) / cos(2<em>θ</em>)
tan(2<em>θ</em>) = (2 sin(<em>θ</em>) cos(<em>θ</em>)) / (2 cos²(<em>θ</em>) - 1)
tan(2<em>θ</em>) = (2 (-3/5) (-4/5)) / (2 (-4/5)² - 1)
tan(2<em>θ</em>) = 24/7
8 because it goes from 1s 10s 100s so 8 is the answer
