It has been proven that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
<h3>How to prove a Line Segment?</h3>
We know that in a triangle if one angle is 90 degrees, then the other angles have to be acute.
Let us take a line l and from point P as shown in the attached file, that is, not on line l, draw two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In ΔPNM, ∠N = 90°
∠P + ∠N + ∠M = 180° (Angle sum property of a triangle)
∠P + ∠M = 90°
Clearly, ∠M is an acute angle.
Thus; ∠M < ∠N
PN < PM (The side opposite to the smaller angle is smaller)
Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to all of them. Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
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Answer:
Equation
28 = X - 12
Solve for X
Isolate the X
28+12 = X
40 = X
Step-by-step explanation:
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Answer:
See below in bold.
Step-by-step explanation:
cos 4x = 2cos^2 2x - 1 = 2 (2 cos^2 x - 1)^2 - 1
and cos 2x = 2 cos^2 x - 1 so we have:
2 ( 2 cos^2 x - 1)^2 - 1 - (2cos^2 x - 1) = 0
2 ( 2 cos^2 x - 1)^2 - 2 cos^2 x = 0
(2 cos^2 x - 1)^2 - cos^2 x = 0
Let c = cos^2 x, then:
(2c - 1)^2 - c = 0
4c^2 - 4c + 1 - c = 0
4c^2 - 5c + 1 = 0
c = 0.25, 1
cos^2 x = 0.25 gives cos x = +/- 0.5
and cos^2 x = 1 gives cos x = +/- 1.
So for x = +/- 1 , x = 0, π.
For cos x = +/- 0.5, x = π/3, 2π/3, 4π3,5π/3.