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March 7th, 2019, 06:36 AM   #1
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Geometry-a-go-go

Code:
A                b                B


a                                  a


D                b                C

                                    c

                                    E
Rectangle ABCD, dimensions a by b.
BC is extended to E, CE = c, angle CED = 75 degrees.
DE = 2a + c.

Angle BDC = ?

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March 7th, 2019, 01:17 PM   #2
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Basic equations: $\frac{b}{c}=\tan(75^\circ)$, $(2a+c)^2=b^2+c^2$. Solve and get $r^2+r\cot(75^\cdot)-\frac{1}{4}=0$, where $r=\frac{a}{b}=\tan(\angle BDC)$.

I'll leave the algebra to others.
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Last edited by skipjack; March 7th, 2019 at 03:26 PM.
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March 7th, 2019, 02:08 PM   #3
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Agree...which results in angleBDC = 20.989984...degrees.

I guess that eliminates a possible non-trig solution:
if result was exactly 21 degrees instead, then a non-trig solution
would at least "have a chance!".
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