Angle G = 130 degrees
Angle H = 50 degrees
Angle K = 74 degrees
Angle M = 106 degrees
<u>Angle G would be 130 degrees</u> because it's a vertical angle, and vertical angles are always alike.
<u>Angle H would be 50 degrees</u> because it's an adjacent angle, and we also know that one side of the line is always 180 degrees so we have an equation that looks like this 180 - 130 = 50 degrees
<u>Angle K would be 74 degrees</u> because it's a vertical angle.
<u>Angle M would be 106 degrees</u> because it's an adjacent angle.
24.9 miles/hour = 24.9/60 miles/minute = .415 miles/minute
<span>distance = speed * time </span>
<span>= .415 * 30 = 12.45 miles</span>
Answer:
all work is shown and pictured
A. You can factor it to those terms
Slope of line = tan(120) = -tan(60) = - √3
Distance from origin = 8
Let equation be Ax+By+C=0
then -A/B=-√3, or
B=A/√3.
Equation becomes
Ax+(A/√3)y+C=0
Knowing that line is 8 units from origin, apply distance formula
8=abs((Ax+(A/√3)y+C)/sqrt(A^2+(A/√3)^2))
Substitute coordinates of origin (x,y)=(0,0) =>
8=abs(C/sqrt(A^2+A^2/3))
Let A=1 (or any other arbitrary finite value)
solve for positive solution of C
8=C/√(4/3) => C=8*2/√3 = (16/3)√3
Therefore one solution is
x+(1/√3)+(16/3)√3=0
or equivalently
√3 x + y + 16 = 0
Check:
slope = -1/√3 .....ok
distance from origin
= (√3 * 0 + 0 + 16)/(sqrt(√3)^2+1^2)
=16/2
=8 ok.
Similarly C=-16 will satisfy the given conditions.
Answer The required equations are
√3 x + y = ± 16
in standard form.
You can conveniently convert to point-slope form if you wish.
