Question

6. If mZCDF = (3x + 14)”, mZFDE = (5x – 2)”, and mZCDE = (10x – 18)”, find each measure.
D
E
XE
F
mZCDF =
mZFDE =
mZCDE =

1. diemthu

x = 15 , m∠ CDF = 59° , m∠FDE = 73° , m∠CDE = 132°

Step-by-step explanation:

* Lets explain how to solve the problem

– The m∠ CDF = (3x + 14)°

– The m∠ FDE = (5x – 2)°

– The m∠ CDE = (10x – 18)°

– We need to find the value of x and the measure of each angle

∵ Point D is the vertex of the three angles

∵ The ray DF is between the rays DC and DE

∴ m∠ CDE = m∠ CDF + m∠ FDE

∵ m∠ CDE = (10x – 18)°

∵ m∠ CDF = (3x + 14)°

∵ m∠ FDE = (5x – 2)°

– Substitute the measures of these angles in the equation above

∴ 10x – 18 = (3x + 14) + (5x – 2)

– Add the like terms in the right hand side

∴ 10x – 18 = (3x + 5x) + (14 – 2)

∴ 10x – 18 = 8x + 12

– Subtract 8x from both sides and add 18 to both sides

∴ 10x – 8x = 12 + 18

∴ 2x = 30

– Divide both sides by 2

∴ x = 15

– Substitute the value of x in the measure of each angle to find it

∴ m∠ CDF = 3(15) + 14 = 59°

∴ m∠ FDE = 5(15) – 2 = 73°

∴ m∠ CDE = 10(15) -18 = 132°

2. thongdat2

9514 1404 393

∠CDF = 59°

∠FDE = 73°

∠CDE = 132°

Step-by-step explanation:

We can use the angle sum theorem to relate the angles:

∠CDF + ∠FDE = ∠CDE

(3x +14) +(5x -2) = (10x -18) . . . . . divide by ” (presumably, degrees (°))

8x +12 = 10x -18 . . . . . collect terms

30 = 2x . . . . . . . . . . . . add 18-8x

15 = x . . . . . . . . . . . . . . divide by 2

Then the angle measures are …

∠CDF = (3x +14)° = (3(15) +14)° = 59°

∠FDE = (5x -2)° = (5(15) – 2)° = 73°

∠CDE = (10x -18)° = (10(15) -18)° = 132°