Answer
$(a)\space B=178\space units$
$(b)\space C=164\space units$
Work Step by Step
Please see the attached image first.
Here we use the component method to find B and C.
Since the resultant vector of these vectors is zero, we can write.
$\vec A+\vec B+\vec C=0$
This means that the x, and y-components of this equation must be zero.
So, we can write,
$A_{x}+B_{x}+C_{x}=0$ ; Let's plug known values into this equation.
$-145\space units\times cos35^{\circ}\space units+Bsin65^{\circ}-Csin15^{\circ}=0$
$-119\space units+B(0.906)-C(0.259)=0$
$B=\frac{(0.259C+119\space units)}{0.906}-(1)$
Similarly,
$A_{y}+B_{y}+C_{y}=0$ ; Let's plug known values into this equation.
$-145\space units\times sin35^{\circ}\space units+Bcos65^{\circ}-Ccos15^{\circ}=0$
$83.2\space units+B(0.423)-C(0.966)=0-(2)$
(b) (1)=>(2)
$83.2\space units+\frac{(0.259C+119\space units)}{0.906}(0.423)-C(0.966)=0$
$83.2(0.906)\space units+50.337\space units=C(0.875-0.11)$
$C=164\space units$
(a) (1)=>
$B=\frac{0.259\times164\space units+119\space units}{0.906}=178\space units$
