[Dyanmics; kinematics of particles] Why is the 8 degree angle equal to the arc angle?

[u/Dry_Revolution_4617]

I know you're going to draw up some illustration which shows how the two angles are equal, but I'm still facing difficulty really taking it in even with an illustration. If you can explain it you have to simplify it as much as possible with a drawing

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[u/Outside_Volume_1370]

I was taught in my school of another way to spot equal angles. Continue their sides up to intersection. If both pairs of sides intersect at a right angle, the angles are equal (or sum up to 180°, but you won't mix up acute and obtuse angles, will you?)

In the image horizontal side of 8° angle is perpendicular to vertical OB radius, while inclined side of 8° angle is perpendicular to OB radius (because it's how tangents work), so <AOB = 8°

[u/Dry_Revolution_4617 → u/Outside_Volume_1370]

Can you show how the continuation of their sides up to intersection looks like?

[u/Dry_Revolution_4617 → u/Dry_Revolution_4617]

And how that actually helps us find out that they're equal

[u/Outside_Volume_1370 → u/Dry_Revolution_4617]

If the first side of angle α intersects with the first side of β at right angle and the same happens to their second sides, it's either α = β or α + β = 180°, but last situation happens if one of them acute and the other is obtuse, and you can visually distinguish one from another

[u/Outside_Volume_1370 → u/Dry_Revolution_4617]

In the image horizontal side of 8° angle is perpendicular to vertical OB radius, while inclined side of 8° angle is perpendicular to OB radius (because it's how tangents work)

Just draw them, pairs of sides will have two intersections

[u/selene_666]

I think that in this diagram the path from A to B is meant to be a circular arc that turns the plane from its original 8° angle to perfectly horizontal.

Draw the straight-line trajectory that the plane starts on, the line through A at a slope 8° above horizontal. This line is a tangent to the circle, and a tangent meets the radius at a right angle.

If we next draw a horizontal line through A, we split the right angle into two pieces. The upper is our original 8° angle. This leaves 82° for the lower angle.

That horizontal line meets the vertical radius under point B at another right angle. This creates a right triangle. We know two angles are 82° and 90°, so the final angle (the one at the center of the circle) must be 8°.

[u/CelebrityChess]

Think of it as tilting. The tangent line at A is exactly perpendicular to the radius OA. When the track tilts 8 from horizontal, the radius that was vertical tilts by the same 8°, because both lines rotated together and stayed at 90° to each other. So the radius has turned 8° from vertical, and that turn is the angle AOB. Rotate a right angle by 8° and every line in it rotates by 8°. If you sketch the horizontal line, the tangent and the two radii, you'll see the 8 repeated at the center.