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asnaseer
 3 years ago
Geometry challenge. Question to follow in drawing.
asnaseer
 3 years ago
Geometry challenge. Question to follow in drawing.

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asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4there are 3 squares next to each other. prove C = A + B

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.13 squares with the same side length I presume?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4yes  3 identical squares

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4BTW: This is a geometry challenge  so no trig allowed :)

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1Okay I can solve it using height and distance, one interesting fact: \( \sin(CB) = \sin (C) \sin (B) \)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4:D  think "outside" the box

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1Comone why not use trig? :P

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4ok  lets see the trig solution. but there is a very elegant geometric solution if you can find it. :)

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm Asnaseer rules :D

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4I'm off to have some food  will be back soon ...

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.1I can't seem to be able to find a geometrical proof. Here's a proof using trigonometry: We have \(\sin C=\frac{1}{\sqrt{2}}\), \(\sin B=\frac{1}{\sqrt{5}}\), \(\sin A=\frac{1}{\sqrt{10}}\), and \(\cos A=\frac{3}{\sqrt{10}}\), \(\cos B=\frac{2}{\sqrt{5}}.\) It's easy to show that \(\sin C=\sin(A+B)\) and hence \(C=A+B \text{ because } A,B, C \in (0, \frac{\pi}{2}).\)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4I'll let this problem /simmer/ for a while before showing the geometric proof  unless of course someone actually proves it by then :)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4I gave a hint above  think "outside" the box :)

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4I guess I can give another hint to you guys, here it is...

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4dw:1328468434372:dw essentially, you want to try and prove that D=B

Shayaan_Mustafa
 3 years ago
Best ResponseYou've already chosen the best response.0It can be proof using Euclidean concepts of geometry i.e. using theorems, postulates, axiom etc... It is actually difficult. But intresting.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.4There is actually a much easier proof using only the properties of similar triangles.

Shayaan_Mustafa
 3 years ago
Best ResponseYou've already chosen the best response.0Yes I know in terms of Euclidean Geometry. These properties of triangles are followed by, called, postulates.
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