3 3 square

Geometry Level 2

The A D E H ADEH rectangle is divided into 3 3 congruent square pieces. Calculate α + β \alpha +\beta .

30 { 30 }^{ { \circ } } 60 { 60 }^{ { \circ } } 90 { 90 }^{ { \circ } } 45 { 45 }^{ { \circ } }

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2 solutions

Yuriy Kazakov
Mar 11, 2018

We see isosceles right triangle B F M BFM with an angle of 4 5 45^{\circ} at the base. And α = M B N \angle \alpha =\angle MBN , β = F B N \angle \beta =\angle FBN and F B N + M B N = 4 5 \angle FBN+\angle MBN=45^{\circ} .

sin ( α + β ) = sin α cos β + cos α sin β sin ( α + β ) = a a 10 2 a a 5 + 3 a a 10 a a 5 sin ( α + β ) = 2 50 + 3 50 sin ( α + β ) = 5 50 sin ( α + β ) = 5 5 2 sin ( α + β ) = 1 2 sin ( α + β ) = 1 2 2 α + β = 45 \large \sin { (\alpha +\beta ) } =\sin { \alpha } \cos { \beta } +\cos { \alpha } \sin { \beta } \\ \large \sin { (\alpha +\beta ) } =\frac { a }{ a\sqrt { 10 } } \cdot \frac { 2a }{ a\sqrt { 5 } } +\frac { 3a }{ a\sqrt { 10 } } \cdot \frac { a }{ a\sqrt { 5 } } \\ \large \sin { (\alpha +\beta ) } =\frac { 2 }{ \sqrt { 50 } } +\frac { 3 }{ \sqrt { 50 } } \\ \large \sin { (\alpha +\beta ) } =\frac { 5 }{ \sqrt { 50 } } \\ \large \sin { (\alpha +\beta ) } =\frac { 5 }{ 5\sqrt { 2 } } \\ \sin { (\alpha +\beta ) } =\frac { 1 }{ \sqrt { 2 } } \\ \sin { (\alpha +\beta ) } =\frac { 1 }{ 2 } \sqrt { 2 } \\ \large \therefore \alpha +\beta ={ 45 }^{ \circ }

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