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6. if (mangle cdf=(3x + 14)^{circ}), (mangle fde=(5x - 2)^{circ}), and …

Question

  1. if (mangle cdf=(3x + 14)^{circ}), (mangle fde=(5x - 2)^{circ}), and (mangle cde=(10x - 18)^{circ}), find each measure.

(mangle cdf=)
(mangle fde=)
(mangle cde=)
(x =)

Explanation:

Step1: Set up equation based on angle - sum

Since $\angle CDF+\angle FDE=\angle CDE$, we have $(3x + 14)+(5x - 2)=10x-18$.

Step2: Simplify left - hand side

Combine like terms: $3x+5x + 14 - 2=8x + 12$. So the equation becomes $8x+12 = 10x-18$.

Step3: Solve for x

Subtract $8x$ from both sides: $12=10x - 8x-18$, which simplifies to $12 = 2x-18$. Then add 18 to both sides: $12 + 18=2x$, so $30 = 2x$. Divide both sides by 2: $x = 15$.

Step4: Find $\angle CDF$

Substitute $x = 15$ into the expression for $\angle CDF$: $m\angle CDF=3x + 14=3\times15+14=45 + 14=59^{\circ}$.

Step5: Find $\angle FDE$

Substitute $x = 15$ into the expression for $\angle FDE$: $m\angle FDE=5x - 2=5\times15-2=75 - 2=73^{\circ}$.

Step6: Find $\angle CDE$

Substitute $x = 15$ into the expression for $\angle CDE$: $m\angle CDE=10x-18=10\times15-18=150 - 18=132^{\circ}$.

Answer:

$x = 15$
$m\angle CDF=59^{\circ}$
$m\angle FDE=73^{\circ}$
$m\angle CDE=132^{\circ}$