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Let's go through each question step-by-step.

### Question 1
You need to find the phase difference between the two voltages:
\[
v_1(t) = -\sin(t - 135^\circ)
\]
and
\[
v_2(t) = 2 \cos(t + 120^\circ).
\]

1. **Convert Both Equations to the Same Trigonometric Function**: It is easier to compare phase angles if both functions use either sine or cosine.
   - We know that \( \cos(t + 120^\circ) = \sin(t + 120^\circ + 90^\circ) = \sin(t + 210^\circ) \).
   - Thus, \( v_2(t) = 2 \sin(t + 210^\circ) \).

2. **Rewrite \( v_1(t) \) in Positive Form**:
   - \( -\sin(t - 135^\circ) = \sin(t - 135^\circ - 180^\circ) = \sin(t - 315^\circ) \).

3. **Compare Phases**:
   - The phase of \( v_1(t) \) is \( -315^\circ \) (or equivalently \( +45^\circ \)).
   - The phase of \( v_2(t) \) is \( +210^\circ \).

4. **Find the Phase Difference**:
   - Phase difference = \( |45^\circ - 210^\circ| = 165^\circ \).

So, the phase difference between \( v_1(t) \) and \( v_2(t) \) is **165 degrees**.

---

### Question 2
Find the resultant of the complex number \(\frac{1 + j2}{2 - j}\) in both Cartesian and Polar forms.

1. **Multiply by the Complex Conjugate**:
   \[
   \frac{1 + j2}{2 - j} \cdot \frac{2 + j}{2 + j} = \frac{(1 + j2)(2 + j)}{(2 - j)(2 + j)}.
   \]

2. **Calculate the Numerator**:
   \[
   (1 + j2)(2 + j) = 1 \cdot 2 + 1 \cdot j + j2 \cdot 2 + j2 \cdot j = 2 + j + j4 - 2 = 4j + j = 

1. Find the phase difference between the following two voltages:

v1(t) = -sin(t - 135°) & v2(t) = 2 cos(t + 120°).

2. Find the resultant of the following complex number in both Cartesian and Polar representations:

(1 + j2) / (2 - j)

Learn how to calculate the phase difference between two sinusoidal voltages and determine the resultant of a complex number in both Cartesian and Polar forms. This solution covers trigonometric identities and complex number manipulation, ideal for engineering students.

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