Question

So,some please help me with this question !

Answer 1

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$\qquad❖ \: \sf \: \angle C = 63 \degree$

$\textsf{ \underline{\underline{Steps to solve the problem} }:}$

$\qquad❖ \: \sf \:x + 4x + 12 + x=180°$

( Angle EBA= Angle DBC, and the three angles sum upto 180° due to linear pair property )

$\qquad❖ \: \sf \:6x + 12 = 180$

$\qquad❖ \: \sf \:6x = 180 - 12$

$\qquad❖ \: \sf \:6x = 168$

$\qquad❖ \: \sf \:x = 28 \degree$

Next,

$\qquad❖ \: \sf \: \angle C + \angle D + x = 180°$

$\qquad❖ \: \sf \: \angle C +3x + 5 + x = 180°$

$\qquad❖ \: \sf \: \angle C +4x = 180 - 5$

$\qquad❖ \: \sf \: \angle C +4(28) = 175$

( x = 28° )

$\qquad❖ \: \sf \: \angle C +112 = 175$

$\qquad❖ \: \sf \: \angle C = 175 - 112$

$\qquad❖ \: \sf \: \angle C = 63 \degree$

$\qquad \large \sf {Conclusion} :$

$\qquad❖ \: \sf \: \angle C = 63 \degree$

Answer 2

Answer:  63 degrees  (choice A)

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Explanation:

Angles EBA and DBC are congruent because of the similar arc marking. Both are x each.

Those angles, along with EBD, combine to form a straight angle of 180 degrees. We consider those angles to be supplementary.

So,

(angleEBA) + (angleEBD) + (angleDBC) = 180

( x ) + (4x+12) + (x) = 180

(x+4x+x) + 12 = 180

6x+12 = 180

6x = 180-12

6x = 168

x = 168/6

x = 28

Angles EBA and DBC are 28 degrees each.

This means angle D = 3x+5 = 3*28+5 = 89

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Then we have one last set of steps to finish things off.

Focus entirely on triangle DBC. The three interior angles add to 180. This is true of any triangle.

D+B+C = 180

89 + 28 + C = 180

117+C = 180

C = 180 - 117

C = 63 degrees