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amm1812
1 year ago
10

Find the values of the missing sides in the parallelogram given.

Mathematics
1 answer:
natta225 [31]1 year ago
6 0

Given:

m∠B = 44°

Let's find the following measures:

m∠A, m∠BCD, m∠CDE

We have:

• m∠A:

Angle A and Angle B are interior angles on same side of a transversal.

The interior angles are supplementary.

Supplementary angles sum up to 180 degrees

Therefore, we have:

m∠A + m∠B = 180

m∠A + 44 = 180

Subtract 44 from both sides:

m∠A + 44 - 44 = 180 - 44

m∠A = 136°

• m,∠,BCD:

m∠BCD = m∠A

Thus, we have:

m∠BCD = 136°

• m∠CDE:

Angle C and angle CDE form a linear pair.

Linear pair of angles are supplementary and supplementary angle sum up to 180 degrees.

Thus, we have:

m∠D = m∠B

m∠D = 44°

m∠CDE + m∠D = 180

m∠CDE + 44 = 180

Subract 44 from both sides:

m∠CDE + 44 - 44 = 180 - 44

m∠CDE = 136°

ANSWER:

• m∠A = 136°

,

•

,

• m∠BCD = 136°

,

•

,

• m∠CDE = 136°

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For the second term:

\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}

So, all together: (same denominator; combine terms)

\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}

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Everything simplifies to 1.

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