Geometry Proof (for fun)
|dw:1502651945890:dw|
Assume that this rectangle is composed of 3 squares, each with equal side length. Prove that A + B + C = 90 degrees Hint: think outside the box
with \(A = {\pi \over 4}\), need to show that \(B + C = {\pi \over 4}\) so tan addition forms: \(\tan B = {1 \over 2}\) \(\tan C = {1 \over 3}\) \(\implies \tan (B +C) = \dfrac{\tan B + \tan C }{1 - \tan B \tan C} = \dfrac{\frac{1}{2} + \frac{1}{3} }{1 - \frac{1}{2} * \frac{1}{3} } = 1\) not the beautiful solution but *a* solution
Good answer There are a lot of ways to solve this, I'll post another solution later tonight
look forward to a cool solution :)
for this solution we draw an identical rectangle above the first rectangle, partitioned in the same way
|dw:1502675191558:dw|
we can draw another triangle like so:|dw:1502675258798:dw|
now if we shade in these two regions:
|dw:1502675297868:dw|
(the graph isn't drawn well but the two black triangles are equal AND they are equal to the triangle containing angle B
|dw:1502675431468:dw|
now if we look at the big triangle in the middle, the legs are equal, making the lower right angle equal to 45
|dw:1502675506363:dw|
without knowing any trig, we can see that the original angle A is 90/2 = 45 since the right angle is bisected
|dw:1502675558025:dw|
therefore A + B + C = the angle in the lower right corner = 90 degrees
I will admit it is more work than the direct calculations but I think it is very clever @sillybilly123
V that is super-smart !!
classy :)
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