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# three squares

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The diagram shows three squares.  Find the sum of the angles alpha, beta, and gamma. May 3, 2020

#3
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There is a rather obscure formula for the sum of two inverse tangents:

tan-1(a) + tan-1(b)  =  tan-1(  (a + b / ( 1 - a·b ) )

tan-1( 1/2 ) + tan-1( 1/3 )  =  tan-1(  ( 1/2 + 1/3 ) / ( 1 - ( 1/2 )·( 1/3 ) )

=  tan-1( (5/6) / (5/6) )

=  tan-1( 1 )

=  pi/4

Adding this to  tan-1( 1 )  gives a total of  pi/2  [or 90o]

May 3, 2020

#1
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sum  =  tan–1(1/1) + tan–1(1/2) + tan–1(1/3)

sum  =  45.000o + 25.565o + 18.435o

sum  =  89.000o

.

May 3, 2020
#2
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alpha     =  tan-1(1/1)

beta       =  tan-1(1/2)

gamma  =  tan-1(1/3)

The sum will be  pi/2.

May 3, 2020
#3
+1

There is a rather obscure formula for the sum of two inverse tangents:

tan-1(a) + tan-1(b)  =  tan-1(  (a + b / ( 1 - a·b ) )

tan-1( 1/2 ) + tan-1( 1/3 )  =  tan-1(  ( 1/2 + 1/3 ) / ( 1 - ( 1/2 )·( 1/3 ) )

=  tan-1( (5/6) / (5/6) )

=  tan-1( 1 )

=  pi/4

Adding this to  tan-1( 1 )  gives a total of  pi/2  [or 90o]

geno3141 May 3, 2020
#4
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Wow!!!....thanks, geno....I've never seen that "formula" before   !!!   CPhill  May 3, 2020