Answer:
[tex]m\angle QSN=65^\circ[/tex]
Step-by-step explanation:
In the given figure, PQRS is a rhombus and SRM is an equilateral triangle.
We are also given that SN⊥RM and that ∠PRS = 55°.
And we want to find the measure of ∠QSN.
Remember that since PQRS is a rhombus, the angles formed by its diagonals are right angles. Let the intersection point of the diagonals be K. Therefore:
[tex]m\angle RKS=90^\circ[/tex]
Now, RKS is also a triangle. The interior angles of all triangles must be 180. Thus:
[tex]m\angle RKS+m\angle KSR+m\angle SRK=180[/tex]
Substitute in known values:
[tex]90+55+m\angle KSR=180[/tex]
Solve for ∠KSR:
[tex]m\angle KSR+145=180\Rightarrow m\angle KSR=35^\circ[/tex]
Since SRM is an equilateral triangle, this means that:
[tex]m\angle SRM=m\angle RMS=m\angle MSR=60^\circ[/tex]
Note that RNS is also a triangle. Therefore:
[tex]m\angle SRM+m\angle RNS+m\angle NSR=180[/tex]
Substitute in known values:
[tex]60+90+m\angle NSR=180[/tex]
So:
[tex]m\angle NSR+150=180\Rightarrow m\angle NSR=30^\circ[/tex]
∠QSN is the addition of the two angles:
[tex]m\angle QSN=m\angle KSR+m\angle NSR[/tex]
Therefore:
[tex]m\angle QSN=35+30=65^\circ[/tex]
