Question

1. if ∠klm = 106, ∠rlm = ______
2. ∠ikh = ______
3. kj = ______

Explanation:

Step1: Solve for x in the first - angle problem

Since $\angle KLM+\angle RLM = 180^{\circ}$ (linear - pair of angles) and $\angle KLM=106^{\circ}$, and $\angle RLM=(2x + 2+6x)^{\circ}$.
First, simplify the expression for $\angle RLM$: $2x + 2+6x=8x + 2$.
Then set up the equation $106+(8x + 2)=180$.
Combine like terms: $8x+108 = 180$.
Subtract 108 from both sides: $8x=180 - 108=72$.
Divide both sides by 8: $x = 9$.
So, $\angle RLM=8x + 2=8\times9+2=74^{\circ}$.

Step2: Solve for x in the second - angle problem

Since $\angle IKH$ is a right - angle (marked with a right - angle symbol), $\angle IKH=(2x - 5+4x - 13)^{\circ}=90^{\circ}$.
Simplify the left - hand side: $2x-5 + 4x-13=6x-18$.
Set up the equation $6x-18 = 90$.
Add 18 to both sides: $6x=90 + 18=108$.
Divide both sides by 6: $x = 18$.
Then find $\angle IKH$:
$\angle IKH=(2x - 5)+(4x - 13)$.
Substitute $x = 18$: $(2\times18-5)+(4\times18-13)=(36 - 5)+(72 - 13)=31 + 59=90^{\circ}$.

Answer:

For $\angle RLM$, the answer is $74^{\circ}$.
For $\angle IKH$, the answer is $90^{\circ}$.