Question

Heya!

Answer 1

Answer:

$m\angle QSN=65^\circ$

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:

$m\angle RKS=90^\circ$

Now, RKS is also a triangle. The interior angles of all triangles must be 180. Thus:

$m\angle RKS+m\angle KSR+m\angle SRK=180$

Substitute in known values:

$90+55+m\angle KSR=180$

Solve for ∠KSR:

$m\angle KSR+145=180\Rightarrow m\angle KSR=35^\circ$

Since SRM is an equilateral triangle, this means that:

$m\angle SRM=m\angle RMS=m\angle MSR=60^\circ$

Note that RNS is also a triangle. Therefore:

$m\angle SRM+m\angle RNS+m\angle NSR=180$

Substitute in known values:

$60+90+m\angle NSR=180$

So:

$m\angle NSR+150=180\Rightarrow m\angle NSR=30^\circ$

∠QSN is the addition of the two angles:

$m\angle QSN=m\angle KSR+m\angle NSR$

Therefore:

$m\angle QSN=35+30=65^\circ$

Answer 2

Answer:

$\displaystyle\sf \angle \: QSN = {65}^{ \circ}$

Step-by-step explanation:

we are given a rombus and an equilateral triangle and $\angle PRS$

said to figure out $\angle QSN$

we know that

every angle of an equilateral triangle is 60°

so

$\displaystyle \angle SRM$ 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 $\displaystyle \angle SXR$ should be 90°

we are also given SN $\displaystyle \perp$ RM

so $\displaystyle\angle RNS$ should be 90°

notice that SXRN is a quadrilateral

so the sum of its interior angles is 360°

according to the question

$\displaystyle\sf\angle RNS + \angle SXR + \angle PRM + \angle \: QSN = 360$

substitute the value of $\angle RNS,\angle SXR\: and \: \angle PRM$

$\displaystyle\sf {90}^{ \circ} + {90}^{ \circ} + {115}^{ \circ} + \angle \: QSN = {360}^{ \circ}$

simplify addition:

$\displaystyle\sf {295}^{ \circ} + \angle \: QSN = {360}^{ \circ}$

cancel 295 from both sides:

$\displaystyle\sf {295}^{ \circ} - {295}^{ \circ} + \angle \: QSN = {360}^{ \circ} - {295}^{ \circ}$

hence

$\displaystyle\sf \angle \: QSN = {65}^{ \circ}$