Answer:
(a) ∠FCB = 70°
(b) ∠DCE = 55°
(c) ∠DCE +∠ECF +∠FCG +∠GCB = 55° +55° +35° +35° = 180°
Explanation:
Given angle bisectors CE and CG, ∠FCG = 35°, ∠DCE = 4x+15, ∠ECF = 6x-5, you want the measures of angles FCB and DCE.
(a) FCB
Ray CG bisects angle FCB, splitting it into two congruent angles. The measure of one of those (∠FCG) is 35°, so the measure of the two of them together is 2×35° = 70°.
∠FCB = 70°.
(b) DCE
Angles DCE and ECF are created congruent by angle bisector CE. This means ...
∠DCE = ∠ECF
4x +15 = 6x -5
20 = 2x
Then angle DCE is ...
4x +15 = 2(2x) +15 = 2(20) +15 = 55 . . . . . . degrees
∠DCE = 55°
(c) Angle Addition
We have already used the Angle Addition theorem to find the measure of ∠FCB:
∠FCB = ∠FCG +∠GCB = 35° +35° = 70°
We could also use the Angle Addition theorem to find ∠DCE:
∠FCD +∠FCB = 180° . . . . . . . . . the sum of the parts gives the whole
2(∠DCE) +2(∠FCG) = 180° . . . . substitute 2 halves for each angle
∠DCE +∠FCG = 90° . . . . . . . . divide by 2
∠DCE = 90° -∠FCG = 90° -35° = 55° . . . . . matches the above calculation
The sum of all of the parts is the whole:
∠DCE +∠ECF +∠FCG +∠GCB = 55° +55° +35° +35° = 180°
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Additional comment
You may notice we did not actually find the value of x in part (b). x=10. We didn't need to know this; we only need to know the value of 4x.
The angle addition theorem is similar to the segment addition theorem. It tells you the whole is the sum of the parts. Along with that, we used the fact that a straight angle has a measure of 180°, or a linear pair (FCD, FCB) are supplementary angles (total 180°).
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