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2 years ago

Please help !!!!!!!!!!!

Mathematics

1 answer:

Dvinal [7]2 years ago

7 0

Answer: 6) x= 17 Same side interior angles sum to 180.

7) x= 3 Alternate exterior angles are equal.

8) x= 35 Corresponding angles are equal.

9) x= 12 Alternate interior angles are equal

Step-by-step explanation:

6) x= Same side interior angles sum to 180.

53+(8x-9)= 180

44+8x= 180

Subtract 44 from both sides

8x= 136

Divide both sides by 8

X = 17

7) x= Alternate exterior angles are equal. 15x+29= 26x-4

Subtract 15x from both sides

29= 11x-4

Add 4 to both sides

33= 11x

Divide both sides by 11

X=3

8) x= Corresponding angles are equal.

4x+7= 6x-63

Subtract 4x from both sides

7 = 2x -63

Add 63 to both sides

70= 2x

Divide both sides by 2

X= 35

9) x= Alternate interior angles are equal

9x+37= 14x-23

Subtract 9 x from both sides

37= 5x -23

Add 23 to both sides

60= 5x

Divide both sides by 5

12=x

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Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

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Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

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$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

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simplify

$sin(\alpha)=\frac{4}{7}$

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$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

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