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m < cxd = m∠gxf = 2x + 15 and m∠gxc = 3x + 5, find m∠cxd.

Question

m < cxd =

m∠gxf = 2x + 15 and m∠gxc = 3x + 5, find m∠cxd.

Explanation:

Step1: Note vertical - angle relationship

Since $\angle GXF$ and $\angle GXC$ are complementary (because $\angle EXB = 90^{\circ}$ and $\angle GXF+\angle GXC+\angle EXB = 180^{\circ}$), we have $\angle GXF+\angle GXC = 90^{\circ}$.
$(2x + 15)+(3x + 5)=90$

Step2: Solve the equation for $x$

Combine like - terms: $2x+3x+15 + 5=90$, so $5x+20 = 90$.
Subtract 20 from both sides: $5x=90 - 20=70$.
Divide both sides by 5: $x=\frac{70}{5}=14$.

Step3: Find $\angle GXC$

Substitute $x = 14$ into the expression for $\angle GXC$: $m\angle GXC=3x + 5=3\times14 + 5=42 + 5=47^{\circ}$.

Step4: Find $\angle CXD$

Since $\angle GXC$ and $\angle CXD$ are complementary (because $\angle GXD = 90^{\circ}$), $m\angle CXD=90^{\circ}-m\angle GXC$.
$m\angle CXD=90 - 47=43^{\circ}$

Answer:

$43$