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Could anyone tell me what this relationship of angles is called and where I can read more about it? I'm not mathematically strong but simply put, if a line is drawn perpendicular to the hypotenuse of a right angled triangle, that angle between that and a vertical line is equal to the angle opposite along the adjacent side to the right angle.

Thanks in advance

Picture demonstrating the above

Xander Henderson
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Gavzooka
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  • Well the B angle is obtained by applying a 90 degrees rotation to the A angle. I do not think that a specific theorem is needed for that. Maybe: Complementary angle theorem. – Frostic Apr 17 '18 at 20:04
  • Thanks, I can see that but I am thinking that there must be a way to explain it and possibly a name for it? – Gavzooka Apr 17 '18 at 20:06
  • It's easy enough to prove. Isn't that an explanation? – saulspatz Apr 17 '18 at 20:08
  • It's an explanation and I know it's true. But mathematically, how do you prove it? What guarantee's that B is equal to A. Perhaps I'm thinking in to it too much. – Gavzooka Apr 17 '18 at 20:10
  • If you translate the B angle so that its intersection matches the intersection of the A angle you obtain alternate exterior angles: https://www.mathsisfun.com/definitions/corresponding-angles.html – Frostic Apr 17 '18 at 20:11
  • Thanks. I see that...I'm still being blinded by something though. I know that's how it works. But if I was to write a formula to prove this, what would it be? – Gavzooka Apr 17 '18 at 20:15

2 Answers2

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Note: The same is true independent if the big triangle has a $90^\circ$ angle or not, or even if it is a triangle. All you need is the tilted line and the horizontal.

Let's have the points $A$ and $B$ where you have the labels for the angles Extend the perpendicular to the tilted line until it intersects the horizontal. Call this $C$. Also, draw the vertical line from $B$. The intersection with $AC$ is $D$. Since angle $B$ and angle $\angle CBD$ are formed by the same lines and are opposite, you have $\angle B=\angle CBD$. The $CBD$ triangle has the angle at $D$ at $90^\circ$, so $\angle BCD=90^\circ-\angle B$. $ABC$ triangle is also a right angle triangle (at $B$), so $\angle A=90^\circ-\angle BCA=90^\circ-(90^\circ-\angle B)=\angle B$.

Andrei
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I don't think it has a name, but the form I heard is this:

  • If two angles have perpendicular sides, they are either equal or 180° complements.
  • If two angles have parallel sides, they are either equal or 180° complements.

I heard it in the context of a physics class since the property is useful for finding angles involving inclined planes, etc.

Note that the angles might be complements if you only know the sides are perpendicular. I found this book: http://www.chegg.com/homework-help/two-angles-respective-sides-perpendicular-angles-either-cong-chapter-4.1-problem-22e-solution-9780321830951-exc

If you would just like to focus on the scenario above, here's a brief proof: Picture

∠ABC = ∠A + 90° (exterior angle property)

∠B = ∠ABC - 90°

∠A = ∠B

Alexander Wu
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  • Thank you. I specifically discovered this when working with an 'Inclined plane'. Thanks for reference. – Gavzooka Apr 17 '18 at 20:47