1) Support reactions

2) Forces on BC, BD by method of section

3) Forces on AB, AE and BE by method of joints

$$M = 2j – R$$

$7 = 2 (5) – 3 \\ 7 = 7 $

LHS = RHS

Hence, perfect truss.

FBD:

In $\triangle AEB$ by pythagoras theorem

$$AB=\sqrt{AE^2+BE^2}=\sqrt{3^2+2^2}=\sqrt {13}$$

$\theta =\tan^{-1}(\dfrac 23)=33.69^\circ \\ \text {similarly,} BD =\sqrt {13} \\ In \triangle BCD , \angle BCD =90^\circ -33.69^\circ =56.31^\circ \\ \text {By cos inerule} \\ BC^2=BD^2 +CD^2 -2\times BD \times CD\times \cos \angle BDC \\ 13+5^2 -2\times \sqrt{13} \times 5\times \cos 56.31 \\ =18 \\ \therefore BC =\sqrt {18}=3\sqrt2 =4.2426 \\ \therefore \cos \beta =\dfrac {BC^2+CD^2-BD^2}{2\times BC\times CD}=\dfrac {18+25-13}{2\times \sqrt{18}\times 5} =\dfrac 1{\sqrt 2} \\ \beta =\cos ^{-1} (\dfrac 1{\sqrt 2}) =45^\circ $

Since the truss is in equilibrium,

$\sum M_A=0 \\ \therefore -100 \times AR=200 \times AE-200 \times AD -150 \times BD +R_B\times AD =0 \\ \therefore R_B \times 6 =100 \times 3+200 \times 6 +150\times 5\\ R_B=475 N\\ Also , \sum F_y=0 \\ R_A\sin a +R_{BG}-100 -200-200=0 \\ R_A \sin a=-25 \rightarrow (1)\\ And ,\sum F_X =0 \\ \therefore R_A \cos a +150=0 \\ R_A \cos a=-150 \rightarrow (2) \\ \text {squaring and adding (1) and (2)} \\ R_A^2\sin ^2a +R_A^2\cos^2a=25^2+(-150)^2 \\ \therefore R_A^2=23125 \\ \therefore R_A=152.0691 N \\ \text { dividing (1) and (2)} \\ \dfrac {R_A\sin a}{R_A \cos a} =\dfrac {25}{-150} \\ \therefore \tan a=-\dfrac 16 \\ \therefore a=\tan^{-1}(\dfrac {-1}6) =170.5377^\circ $

$\sum M_D=0 \\ \therefore F_{CB} \sin \beta \times CD -150 \times CD =0 \\ \therefore F_{CB}\sin 45 \times CD -150 \times CD = 0 \\ \therefore F_{CB} =\dfrac {150}{\sin 45} =212.1320 N \\ Also , \sum F_Y=0 \\ \therefore R_B+F_{DB}\sin \theta -F_{CB}\cos \beta -200 =0 \\ \therefore 475 + F_{DB}\sin 33.69-212.1320 \cos 45-200= 0 \\ \therefore F_{DB} =\dfrac {212.1320\cos 45 +200-475}{\sin 33.69}=225.3474 N \\ And .\sum F_X =0 \\ \therefore -F_{DB}+F_{DB}\cos \theta - F_{CB} \sin \beta +150=0 \\ \therefore 225.3474 \cos 33.69 -212.1320 \sin 45 +150=F_{DB} \\ \therefore F_{DB}=187.5 N \\ \text {we use method of joints :-} \\ \text {Joint A : } \\ \sum F_Y=0 \\ \therefore F_{AB}\sin \theta +R_A \sin a=0 \\ \therefore F_{AB} \sin 33.69 +25=0 (from (1)) \\ \therefore F_{AB} =\dfrac {-25}{\sin 33.69}=-45.0695 N \\ Also , \sum F_X=0 \\ \therefore F_{AB}+R_A\cos a+F_{AB}\cos \theta= 0 \\ \therefore F_{AB} -150 -45.0695 \cos 33.69 =0 (from(2)) \\ \therefore F_{AB}=-187.5 N \\ \text {Joint E :-} \\ \sum F_Y=0 \\ \therefore F_{EB}-200 =0 \\ \therefore F_{EB} =200N$