Answer
The 2nd post is 22.5m from our position.
Work Step by Step
Let North be along the (+)ve y-direction and East be along the (+)ve x-direction.
let
$A$=displacement vector of the 1st post from our position=$52.0m$, $37.0^{\circ}$North of East
$B$=displacement vector from the 1st post to the 2nd post=$68.0m$
$C$=displacement vector of the 2nd post from our position(due South)
Now, using vector addition,we get
$C=A+B$
$C_{x}=A_{x}+B_{x}$ and $C_{y}=A_{y}+B_{y}$->(1)
$=>Ccos(-90^{\circ})=52.0cos(37.0^{\circ})+B_{x}$
$=>B_{x}=-41.53m$
We know , $B_{y}=+-\sqrt (B^{2}-B_{x}^{2})=+-\sqrt (68.0^2-(-41.53)^2)=-53.84m$ [we use (-)ve sign because the 2nd post is due South ]
From equation (1),we get
$Csin(-90^{\circ})=52.0sin(37.0^{\circ})-53.84$
$C=22.5m$
Therefor ,the 2nd post is 22.5m from our position.
