Here, $$\alpha = 5^{\circ}=0.0873\,rad,\, M = 2.5$$ From linearized theory, for a flat plate at an angle of attack $$c_{l}=\frac{4\alpha }{\sqrt{M_{\infty}^{2}-1}}$$ Therefore, $$c_{l}=\frac{4\left ( 0.0873 \right )}{\sqrt{\left ( 2.5 \right )^{2}-1}}=0.1524$$ $$c_{d}=\frac{4\alpha ^{2}}{\sqrt{M_{\infty}^{2}-1}}$$ Therefore, $$c_{d}=\frac{4\left ( 0.0873 \right )^{2}}{\sqrt{\left ( 2.5 \right )^{2}-1}}=0.0133$$ $$C_{p}=\frac{2\theta}{\sqrt{M_{\infty}^{2}-1}}$$ Therefore, $$C_{p}=\frac{2\left ( 0.0873 \right )}{\sqrt{\left ( 2.5 \right )^{2}-1}}=0.0762$$