1. Draw and number a neat, consistent, labeled set of free body diagrams for the parts of the shears.

Note link DE is a two force member with a slope of $m = \dfrac{\var{d_7}} {\var{d_9 - d_4}}$, in tension.  The free body diagrams must be consistent with each other and internal forces must occur in action-reaction pairs.

2. Find the necessary angle.

Let $\theta$ be the angle that DE makes with the horizontal.

$\theta = \tan^{-1} \dfrac{\var{d_7}}{\var{d_9-d_4}} = \var{siground(theta,4)}°$

3. Use FBD II  and take moments about C to find the tension in DE.

$\begin{align}\textrm{II: }\Sigma M_C &= 0 \\ DE_x (\var{d_6}) + DE_y(\var{d_3}) &= P \,(\var{d_3 + d_4 })\\ DE &= \left ( \dfrac{\var{scalar(d_3 + d_4 )}}{\var{scalar(d_6)} \cos \theta + \var{scalar(d_3)}\sin \theta} \right)  P  \\&= \left(\dfrac{\var{d_3 + d_4 }}{\var{siground(dp_1,4)}} \right )P \\&= \var{siground(ma_1,4)} \,P \textrm{ (tension) } \end{align}$

4. FBD III and take moments about B to find the cutting force at A.

$\begin{align}\textrm{III: }\Sigma M_B &= 0 \\ A (\var{d_1}) &=DE_x (\var{d_5 +d_6 +d_7 }) + DE_y(\var{d_8-d_1}) \\ A &= \left ( \dfrac{\var{scalar(d_5 +d_6 +d_7 )}\cos \theta + \var{scalar(d_8-d_1)}\sin \theta}{\var{scalar(d_1 )}} \right) DE \\&= \left(\dfrac{\var{siground(dp_2,4)}}{\var{d_1 }} \right )DE \\&= \var{siground(ma_2,4)} \,DE \\ &= \var{siground(ma_2,4)}\, ( \var{siground(ma_1,4)} \,P )\\ &= \var{siground(ma_t,4)} P \end{align}$

5.  Find the mechanical advantage

Mechanical Advantage =  $\dfrac{A}{P} = \var{siground(ma_t,4)}$

6. Find the force P that must be applied to the handles in order to create a 170 N shearing force at A.

$A = \var{siground(ma_t,4)} P$, so for $A$ = 170 N:

$P = \dfrac{\var{given[1]}}{\var{siground(ma_t,4)}} = \var{siground(answer[1],4)}$