Question

if m∠cdf=(3x + 14)°, m∠fde=(5x - 2)° and m∠cde=(10x - 18)°, find x

Explanation:

Step1: Use angle - addition postulate

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

Step2: Simplify the left - hand side

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

Step3: Isolate the variable terms

Subtract $8x$ from both sides: $8x+12-8x=10x - 18-8x$, which simplifies to $12 = 2x-18$.

Step4: Solve for $x$

Add 18 to both sides: $12 + 18=2x-18 + 18$, getting $30 = 2x$. Then divide both sides by 2: $\frac{30}{2}=\frac{2x}{2}$, so $x = 15$.

Answer:

$15$