$\text{Step 1: Given Data}$

We are tasked with determining the forces in each member of the truss and identifying whether these forces are in tension $T$ or compression $C$.

$$\begin{array}{c}\text{Forces:}\\ P_1=45\text{kN}\\ P_2=30\text{kN}\end{array}$$

$\text{Step 2: Joint C Analysis}$

The free-body diagram of joint C is shown below:

$$\begin{array}{c}\text{Equilibrium in x-direction:}\\ \sum F_x=0\\ -F_{BC}=0\\ F_{BC}=0\text{kN}\end{array}$$

$$\begin{array}{c}\text{Equilibrium in y-direction:}\\ \sum F_y=0\\ F_{CD}–P_1=0\\ F_{CD}=P_1=45\text{kN (T)}\end{array}$$

$\text{Step 3: Joint D Analysis}$

The free-body diagram of joint D is shown below:

$$\begin{array}{c}\text{Equilibrium in y-direction:}\\ \sum F_y=0\\ -F_{CD}–F_{BD}\times \frac{3}{5}=0\\ F_{BD}=-F_{CD}\times \frac{5}{3}\\ =-45\times \frac{5}{3}\\ =-75\text{kN}\\ =75\text{kN (C)}\end{array}$$

$$\begin{array}{c}\text{Equilibrium in x-direction:}\\ \sum F_x=0\\ P_2–F_{AD}–F_{BD}\times \frac{4}{5}=0\\ F_{AD}=P_2–F_{BD}\times \frac{4}{5}\\ =30–(-75)\times \frac{4}{5}\\ =90\text{kN (T)}\end{array}$$

$\text{Conclusion}$

The forces in each member are:
$$\begin{array}{ll}\text{Member BC:}& 0\text{kN}\\ \text{Member CD:}& 45\text{kN in tension (T)}\\ \text{Member BD:}& 75\text{kN in compression (C)}\\ \text{Member AD:}& 90\text{kN in tension (T)}\end{array}$$