1. Given that \(DJ \parallel KE\), \(\angle DBF = 37^\circ\), and \angle \(ABC = 46^\circ\), find \(\angle GCE\) (in degrees).
2. In the diagram, \(\angle CBD = \angle CDB\),\(\angle ABP = 5x-7^\circ\), and \(\angle EDQ = 2x+11^\circ\). What is \(\angle ABD\)(in degrees)?
∠GCE = ∠CBJ because they are corresponding angles.
And
∠CBJ and ∠DBF are vertical angles, so ∠CBJ = ∠DBF
∠DBF = 37º
So... ∠GCE = ∠CBJ = ∠DBF = 37º
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∠ABP and ∠CBD are vertical angles, so ∠ABP = ∠CBD
From the information given, ∠CBD = ∠CDB
∠CDB and ∠EDQ are vertical angles, so ∠CDB = ∠EDQ
So.... ∠ABP = ∠CBD = ∠CDB = ∠EDQ
∠ABP = ∠EDQ
5x - 7 = 2x + 11
3x = 18
x = 6
So we know... ∠ABP = 5x - 7 = 5(6) - 7 = 23°
∠ABD and ∠ABP form a straight line, so ∠ABD + ∠ABP = 180°
∠ABD + 23° = 180°
∠ABD = 180° - 23° = 157°
∠GCE = ∠CBJ because they are corresponding angles.
And
∠CBJ and ∠DBF are vertical angles, so ∠CBJ = ∠DBF
∠DBF = 37º
So... ∠GCE = ∠CBJ = ∠DBF = 37º
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∠ABP and ∠CBD are vertical angles, so ∠ABP = ∠CBD
From the information given, ∠CBD = ∠CDB
∠CDB and ∠EDQ are vertical angles, so ∠CDB = ∠EDQ
So.... ∠ABP = ∠CBD = ∠CDB = ∠EDQ
∠ABP = ∠EDQ
5x - 7 = 2x + 11
3x = 18
x = 6
So we know... ∠ABP = 5x - 7 = 5(6) - 7 = 23°
∠ABD and ∠ABP form a straight line, so ∠ABD + ∠ABP = 180°
∠ABD + 23° = 180°
∠ABD = 180° - 23° = 157°