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# Given that \$DJ \parallel KE\$, \$\angle DBF = 37^\circ\$, and \$\angle ABC = 46^\circ\$, find \$\angle GCE\$ (in degrees). [asy] pair A, B, C, D,

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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)?

Guest Oct 24, 2017
edited by Guest  Oct 24, 2017

#2
+5237
+1

∠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°

hectictar  Oct 24, 2017
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#1
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Only need help with 1 now

Guest Oct 24, 2017
#2
+5237
+1

∠GCE  =  ∠CBJ      because they are corresponding angles.

And

∠CBJ  and  ∠DBF  are vertical angles, so  ∠CBJ  =  ∠DBF

∠DBF  =  37º

So...     ∠GCE  =  ∠CBJ  =  ∠DBF  =  37º

----------

∠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°

hectictar  Oct 24, 2017
#3
+1

Thanks!!!

Guest Oct 24, 2017

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