The question is: Determine the force developed in the member HC of the below truss  (hint use the method of the section and notice member BH makes a right angle at joint B with respect to members AB and BC).

Method of Sections.

F_HC can be determined directly by writing the moment equations of equilibrium about points F and C, respectively, by referring to the FBD of the upper portion of the truss section through 1-1  as shown in Figure below.

$\sum _{}$Fx=0

$\sum _{}$Fy=0

$\sum _{}$M_H =0

Calculating $\angle$BAH : 

tan$\theta$=(1.5/2)

$\theta$=36.87 $°$

and  $\angle$CHG will be :

tan$\varnothing$=(1.5/1)

$\varnothing$=56.3 $°$