Geometric Proof: Finding Angles in an Isosceles Triangle

  • Thread starter Thread starter Sirsh
  • Start date Start date
  • Tags Tags
    Geometric Proof
AI Thread Summary
The discussion focuses on finding angles in an isosceles triangle, specifically within triangle PSR, which contains two isosceles triangles, PSQ and QSR. The initial contributor identifies that angles PSQ and QSR sum to 90 degrees and attempts to establish relationships between the angles using triangle congruence. Another participant clarifies the approach, suggesting to first identify the equal angles in each isosceles triangle. They emphasize that the sum of the angles in triangle PSR must equal 180 degrees. This methodical breakdown aids in solving the angle relationships within the triangle.
Sirsh
Messages
262
Reaction score
10
Hello all, the picture i have attached is the question.
http://img842.imageshack.us/img842/8921/geometricproof.png


I've concluded that there are two i isosceles triangles in this one triangle.

\anglePSQ + \angleQSR = 90degrees
Finding the angle in one of the isosceles triangles.

\DeltaRQS with RQ = SQ
RQ = SQ (given)
If we put a line through the isosceles triangle and call this Z.
RZ = SZ
RZ = SZ (common to both)
therefore: \DeltaRZQ =~ \DeltaSZQ
therefore: \DeltaQRZ = \DeltaQSZ
which results in: \angleQRS = \angleQSR
 
Last edited by a moderator:
Physics news on Phys.org
Hi,
I don't understand your explanation on 'Z'!
But your starting point is correct..You have two isosceles triangles (PSQ and QSR) in the triangle PSR.
I will now give you a easy hint:
First consider the isosceles triangle PSQ and find which two angles are equal?
Then take the second isosceles triangle QSR and find which two angles are equal?
Now you know the three angles of the triangle PSR = 180 degree.
Got it!
 
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...
Back
Top