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tatiyna
3 years ago
15

Quadrilateral MNOP is a rhombus. If the measure of Angle PON = 124, find the measure of Angle POM.

Mathematics
2 answers:
Degger [83]3 years ago
8 0
In a parallelogram, two angles are equal and another two angles are equal. The total amount of angles is 360.124 * 2 = 248. 112 / 2 is 56. The answer is 56.
sergij07 [2.7K]3 years ago
5 0
The answer is 62 because angle POM is the angle bisector of angle PON and an angle bisector is when an angle is divided by 2 because angle POM and MON are congruent. When two angle are congruent  that means that they are both the same number of degrees.
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mars1129 [50]

Answer:

segment FE over segment XY equals segment EG over segment YZ equals segment GF over segment ZX

Step-by-step explanation:

see the attached figure to better understand the problem  

we know that  

If m∠E = m∠Y and m∠F = m∠X  

then  

Triangles EFG and YXZ are similar by AA Similarity Theorem  

Remember that  

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent  

In this problem  

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EG and YZ  

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segment FE over segment XY equals segment EG over segment YZ equals segment GF over segment ZX Firstly , let us learn about trigonometry in mathematics.  

Suppose the ΔABC is a right triangle and ∠A is 90°.  

sin ∠A = opposite / hypotenuse

cos ∠A = adjacent / hypotenuse

tan ∠A = opposite / adjacent

Let us now tackle the problem!  

A similar triangle has the same angle, in other words the triangle has the same shape but different sizes.  

From the figure in the attachment , we can conclude that:  

m∠E = m∠Y  

m∠F= m∠X  

m∠G = m∠Z  

∴ ΔEFG ~ ΔYXZ  ( ΔEFG is similar to ΔYXZ )    

Because of the similarity , then:  

FE : XY = EG : YZ = GF : ZX  

Conclusion:

ΔFG is similar to ΔYXZ.  

Segment FE over segment XY equals segment EG over segment YZ equals segment GF over segment ZX , i.e:    

Keywords: Sine , Cosine , Tangent , Opposite , Adjacent , Hypotenuse , Triangle , Fraction , Lowest , Function , Angle

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Gre4nikov [31]
<h2>Hello!</h2>

The answer is:

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