When two parallel lines are intersected by a transversal, the same-side exterior angles are supplementary. That means that their sum is 180.
Using that logic, if the two roads were parallel, then the sum of their same-side exterior angles will add up to 180. Yet their same-side exterior angles add up to 170 (130 + 40 = 170), hence they can't be parallel.
See the drawing attached below.
Using supplmenatry angles (two angles whose sum of measures add up to 180 or a straight line), we can say that:
m<DIE + m<HID = 18
40 + m<HID = 180
m<HID = 140
Similarly:
m<BHC + m<CHI = 180
130 + m<CHI = 180
m<CHI = 50
Using verticle angles therome, (when two lines intersect, the angles opposite to eachother are congruent, or have the same measure), we can say that:
m<DIE = m<GIH = 40
m<GIE = m<HID = 140
m<CHI = m<AHB = 50
m<BHC = m<AHI = 130
Answer: Does not exist.
Step-by-step explanation:
Since, given function, f(x) = 6x tan x, where −π/2 < x < π/2.
⇒ f(x) = 
And, for vertical asymptote, cosx= 0
⇒ x = π/2 + nπ where n is any integer.
But, for any n x is does not exist in the interval ( -π/2, π/2)
Therefore, vertical asymptote of f(x) where −π/2 < x < π/2 does not exist.
The two posts are parallel lines. The ground is a transversal.
The angles measuring 4x and 75 are same side interior angles, so they are supplementary. Their measures add up to 180.
4x + 75 = 180
4x = 105
x = 26.25