Sending end active power from ABCD, 132/130.5 kV

Use the sending-end power-circle equation (taking $D = A$). With line voltages in kV, the result is in three-phase MW:

$$P_S = \frac{|A||V_S|^2}{|B|}\cos(\beta-\alpha) - \frac{|V_S||V_R|}{|B|}\cos(\beta+\delta)$$

Here $\delta$ is the angle by which $V_S$ leads $V_R$. Since $V_S = 132\angle0^\circ$ and $V_R = 130.5\angle10^\circ$,

$$\delta = 0^\circ - 10^\circ = -10^\circ$$

Data: $|A| = 0.97$, $\alpha = 0.5^\circ$, $|B| = 100\ \Omega$, $\beta = 75^\circ$.

Substitution

$$\begin{aligned} \frac{|A||V_S|^2}{|B|} &= \frac{0.97\times132^2}{100} = 169.01 \\ \frac{|V_S||V_R|}{|B|} &= \frac{132\times130.5}{100} = 172.26 \\ \beta-\alpha &= 74.5^\circ,\quad \beta+\delta = 65^\circ \end{aligned}$$ $$\begin{aligned} P_S &= 169.01\cos74.5^\circ - 172.26\cos65^\circ \\ &= 169.01(0.2672) - 172.26(0.4226) \\ &= 45.17 - 72.80 = -27.63\ \text{MW} \end{aligned}$$

Meaning

$P_S$ is negative because the receiving-end voltage leads the sending-end voltage by $10^\circ$. Active power flows from the leading to the lagging end, so about 27.6 MW actually flows out of the line at the sending end (from the "receiving" bus towards the "sending" bus).

Answer: $P_S$ ≈ −27.63 MW (27.63 MW flows into the sending-end bus).