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

Mr.MathBest ResponseYou've already chosen the best response.1
3 squares with the same side length I presume?
 2 years ago

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

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

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

asnaseerBest ResponseYou've already chosen the best response.4
:D  think "outside" the box
 2 years ago

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

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

FoolForMathBest ResponseYou've already chosen the best response.1
Hmm Asnaseer rules :D
 2 years ago

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

Mr.MathBest ResponseYou've already chosen the best response.1
I 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 \(C=A+B \text{ because } A,B, C \in (0, \frac{\pi}{2}).\)
 2 years ago

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

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

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

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

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

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

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