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!
![Heya![tex] \star \underline{ \underline{ \large{ \text { {Question}}}}} : [/tex] In The Given Figure](https://brainimage.s3.us-east-005.backblazeb2.com/contents/attachments/ULMeU0NSHYj4MHyMn4SnMeHiXWLz8cxQ.jpeg)
Answers
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]
![Heya![tex] \star \underline{ \underline{ \large{ \text { {Question}}}}} : [/tex] In The Given Figure](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/U3HZFIHT9ZX8ofZvDXKhUEIv09UJC4Oo.jpeg)
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