Question

3
$(10x + 2)^{circ}$
g
$(12x - 4)^{circ}$
f
$x=$
$\angle dge=$

Explanation:

Step1: Use the Angle - Bisector Theorem (HL congruence implies equal angles)

Since \(DE = FE\) (marked as equal) and \(GE\) is common, and \(\angle GDE=\angle GFE = 90^{\circ}\), by the Hypotenuse - Leg (HL) congruence criterion for right - triangles \(\triangle GDE\cong\triangle GFE\). Then \(\angle DGE=\angle FGE\). So \(10x + 2=12x-4\).

Step2: Solve the equation for \(x\)

Subtract \(10x\) from both sides of the equation \(10x + 2=12x-4\):
\(2=12x-10x - 4\)
\(2 = 2x-4\)
Add \(4\) to both sides: \(2 + 4=2x\), so \(6 = 2x\).
Divide both sides by \(2\): \(x=\frac{6}{2}=3\).

Step3: Find \(\angle DGE\)

Substitute \(x = 3\) into the expression for \(\angle DGE=(10x + 2)^{\circ}\).
\(\angle DGE=(10\times3 + 2)^{\circ}=(30 + 2)^{\circ}=32^{\circ}\).

Answer:

\(x = 3\); \(\angle DGE=32^{\circ}\)