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In the square ABCD shown above, what is the measure of an yellow angle ?
A. \(15°\)
B. \(20°\)
C. \(25°\)
D. \(30°\)
E. \(45°\)
22 Nov 2021, 05:50
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Official Solution:
In the square ABCD shown above, what is the measure of an yellow angle ?
A. \(15°\)
B. \(20°\)
C. \(25°\)
D. \(30°\)
E. \(45°\)
Construct triangle BFC congruent to triangle DEC on side BC (simply put draw triangle BFC on BC the same way DEC is drawn on DC). Notice that \(\angle BFC =180°-(15°+15°)=150°\).
Since BFC and DEC are congruent, then \(CE = CF\).
Since \(\angle DCE = \angle BCF = 15°\), then \(\angle ECF = 90°-(\angle DCE + \angle BCF) = 60°\)
Thus, triangle EFC is isosceles (\(CE = CF\)) and \(\angle ECF =60°\), which means that triangle EFC is equilateral and all its angles are \(60°\).
Next, since the sum of the angles around point F must be \(360°\), then \(\angle BFE = 360° - (\angle BFC + \angle CFE)=360° - (150° + 60°)=150°\)
Now, consider triangles EFB and CFB. In these triangles, two sides and angle between them are equal (BF is shared side, FC = FE because EFC is equilateral, and \(\angle BFE = \angle BFC=150°\), thus EFB and CFB are congruent.
Finally, since EFB and CFB are congruent, then \(\angle FBC=\angle FBE=15°\) and therefore, \(\angle CBE = \angle FBC+\angle FBE=30°\)
Answer: D
In the square ABCD shown above, what is the measure of an yellow angle ?
A. \(15°\)
B. \(20°\)
C. \(25°\)
D. \(30°\)
E. \(45°\)
Construct triangle BFC congruent to triangle DEC on side BC (simply put draw triangle BFC on BC the same way DEC is drawn on DC). Notice that \(\angle BFC =180°-(15°+15°)=150°\).
Since BFC and DEC are congruent, then \(CE = CF\).
Since \(\angle DCE = \angle BCF = 15°\), then \(\angle ECF = 90°-(\angle DCE + \angle BCF) = 60°\)
Thus, triangle EFC is isosceles (\(CE = CF\)) and \(\angle ECF =60°\), which means that triangle EFC is equilateral and all its angles are \(60°\).
Next, since the sum of the angles around point F must be \(360°\), then \(\angle BFE = 360° - (\angle BFC + \angle CFE)=360° - (150° + 60°)=150°\)
Now, consider triangles EFB and CFB. In these triangles, two sides and angle between them are equal (BF is shared side, FC = FE because EFC is equilateral, and \(\angle BFE = \angle BFC=150°\), thus EFB and CFB are congruent.
Finally, since EFB and CFB are congruent, then \(\angle FBC=\angle FBE=15°\) and therefore, \(\angle CBE = \angle FBC+\angle FBE=30°\)
Answer: D
