Understanding Angle Proofs: How to Identify and Solve for Equal Angles

  • Thread starter Thread starter Miike012
  • Start date Start date
  • Tags Tags
    Angle
AI Thread Summary
The discussion focuses on understanding why two angles in right triangles are equal. The user seeks clarification on the equality of two red angles in their geometry problem. It is explained that the angles are opposite angles, making them equal. Additionally, both triangles share a right angle and another common angle, leading to the conclusion that the remaining angles must also be equal. This reinforces the concept that in triangles with shared angles, the remaining angles will be equal as well.
Miike012
Messages
1,009
Reaction score
0
Quick angle question...

I am going over the sum of two angles proof... and I am confused about one thing..( my geometry isn't that great) can some one tell me how to two red angles are the same...? thank you.
I posted a pick ...
 

Attachments

  • MATH.JPG
    MATH.JPG
    12.6 KB · Views: 424
Physics news on Phys.org


The two triangles the angles are in are right triangles, right?
At the intersection of the vertical line and the lower of the two slanted lines coming out of the origin, there's one angle of the upper right triangle, and another angle of the lower right triangle. These angles are opposite angles, which means they are equal. So we have two triangles, both with a 90°, and each one has an angle that's equal to an angle in the other triangle. That means that the remaining angles of the two triangles (the ones you're asking about) have to be equal as well.
 


Miike012 said:
I am going over the sum of two angles proof... and I am confused about one thing..( my geometry isn't that great) can some one tell me how to two red angles are the same...? thank you.
I posted a pick ...

Call the lower left vertex A. Let the right angle to the right be B, and the angle above C. Make the red angle above that D and the other right angle E.

Angle BCA is the same as DCE (they are reflex angles). Triangle ABC and DCE are both right angles. That gives two triangles with two angles in common, so the third angles (the red angles) must be equal.

(Sorry about the duplication. In the time I tried to post a lettered diagram, Mark 44 got the answer in.)
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Back
Top