Heya!
[tex] \star \underline{ \underline{ \large{ \text { {Question}}}}} : [/tex] In the given figure , PQRS is a rhombus and SRM is an equilateral triangle. If SN [tex] \perp[/tex] RM and [tex] \angle[/tex] PRS = 55° , find the size of [tex] \angle[/tex] QSN.

~Thanks in advance!

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]

Answer:

[tex] \displaystyle\sf \angle \: QSN = {65}^{ \circ} [/tex]

Step-by-step explanation:

we are given a rombus and an equilateral triangle and [tex]\angle PRS[/tex]

said to figure out [tex]\angle QSN[/tex]

we know that

every angle of an equilateral triangle is 60°

so

[tex]\displaystyle \angle SRM[/tex] should be 60°

the point where two diagonals of triangle intercept be x

recall the diagonals of a rhombus intercept each other at right angles

so [tex]\displaystyle \angle SXR [/tex] should be 90°

we are also given SN [tex] \displaystyle \perp[/tex] RM

so [tex]\displaystyle\angle RNS[/tex] should be 90°

notice that SXRN is a quadrilateral

so the sum of its interior angles is 360°

according to the question

[tex] \displaystyle\sf\angle RNS + \angle SXR + \angle PRM + \angle \: QSN = 360[/tex]

substitute the value of [tex]\angle RNS,\angle SXR\: and \: \angle PRM[/tex]

[tex] \displaystyle\sf {90}^{ \circ} + {90}^{ \circ} + {115}^{ \circ} + \angle \: QSN = {360}^{ \circ} [/tex]

simplify addition:

[tex] \displaystyle\sf {295}^{ \circ} + \angle \: QSN = {360}^{ \circ} [/tex]

cancel 295 from both sides:

[tex] \displaystyle\sf {295}^{ \circ} - {295}^{ \circ} + \angle \: QSN = {360}^{ \circ} - {295}^{ \circ} [/tex]

hence

[tex] \displaystyle\sf \angle \: QSN = {65}^{ \circ} [/tex]

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