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iren2701 [21]
3 years ago
9

Find QN. Please show all the work on how you got your answer

Mathematics
1 answer:
Effectus [21]3 years ago
3 0

QN = 28

Solution:

Given MNPQ is a parallelogram.

QT = 4x + 6 and TN = 5x + 4

To find the length of QN:

Let us solve it using the property of parallelogram.

Property of Parallelogram:

Diagonals of the parallelogram bisect each other.

Therefore, QT = TN

⇒ 4x + 6 = 5x + 4

Arrange like terms together.

⇒ 6 – 4 = 5x – 4x

⇒ 2 = x

⇒ x = 2

Substitute x = 2 in QT and TN

QT = 4(2) + 6 = 14

TN = 5(2) + 4 = 14

QN = QT + TN

      = 14 + 14

QN = 28

The length of QN is 28.

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Measure the lengths of the sides of ∆ABC in GeoGebra, and compute the sine and the cosine of ∠A and ∠B. Verify your calculations
marusya05 [52]

Answer:

Sin \angle A =0.80

Cos \angle A=0.60

Sin \angle B =0.60

Cos \angle B=0.80

Step-by-step explanation:

Given

I will answer this question using the attached triangle

Solving (a): Sine and Cosine A

In trigonometry:

Sin \theta =\frac{Opposite}{Hypotenuse} and

Cos \theta =\frac{Adjacent}{Hypotenuse}

So:

Sin \angle A =\frac{BC}{BA}

Substitute values for BC and BA

Sin \angle A =\frac{8cm}{10cm}

Sin \angle A =\frac{8}{10}

Sin \angle A =0.80

Cos \angle A=\frac{AC}{BA}

Substitute values for AC and BA

Cos \angle A=\frac{6cm}{10cm}

Cos \angle A=\frac{6}{10}

Cos \angle A=0.60

Solving (b): Sine and Cosine B

In trigonometry:

Sin \theta =\frac{Opposite}{Hypotenuse} and

Cos \theta =\frac{Adjacent}{Hypotenuse}

So:

Sin \angle B =\frac{AC}{BA}

Substitute values for AC and BA

Sin \angle B =\frac{6cm}{10cm}

Sin \angle B =\frac{6}{10}

Sin \angle B =0.60

Cos \angle B=\frac{BC}{BA}

Substitute values for BC and BA

Cos \angle B=\frac{8cm}{10cm}

Cos \angle B=\frac{8}{10}

Cos \angle B=0.80

Using a calculator:

A = 53^{\circ}

So:

Sin(53^{\circ}) =0.7986

Sin(53^{\circ}) =0.80 -- approximated

Cos(53^{\circ}) = 0.6018

Cos(53^{\circ}) = 0.60 -- approximated

B = 37^{\circ}

So:

Sin(37^{\circ}) = 0.6018

Sin(37^{\circ}) = 0.60 --- approximated

Cos(37^{\circ}) = 0.7986

Cos(37^{\circ}) = 0.80 --- approximated

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3 years ago
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