Question

1. What is the value of x in the triangle?

Answer 1

Answer: Took the test correct :)

Step-by-step explanation:Problem 1

cos(angle) = adjacent/hypotenuse

cos(18) = x/25

25*cos(18) = x

x = 25*cos(18)

x = 23.7764129073789

x = 23.78

Answer: x = 23.78

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Problem 2

theta = greek letter that looks like a "zero" or the letter O with a horizontal line through it

With the reference angle of theta, the opposite side is 3. This side is the leg furthest from the angle theta. In contrast, the side 4 is the closest leg to theta. The longest side 5 is the hypotenuse. The hypotenuse is always the lonest side

sin(theta) = opposite/hypotenuse

sin(theta) = 3/5

cos(theta) = adjacent/hypotenuse

cos(theta) = 4/5

tan(theta) = opposite/adjacent

tan(theta) = 3/4

---------------------------

In summary:

sin(theta) = 3/5

cos(theta) = 4/5

tan(theta) = 3/4

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Problem 3

cos(angle) = adjacent/hypotenuse

cos(x) = 15/29

x = arccos(15/29) .... inverse cosine or cos^(-1) also works

x = 58.852610078491

x = 58.85

Answer: x = 58.85

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Problem 4

The rule we'll use is

sin(x) = cos(90-x)

sin(theta) = cos(90-theta)

cos(90-theta) = cos(58)

90 - theta = 58

90 - theta+theta = 58+theta

90 = 58+theta

90-58 = 58+theta-58

32 = theta

theta = 32

Answer: 32

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Problem 5

x = height of pole

the vertical side is the side we want, which is the opposite side compared to the angle 55 degrees

sin(angle) = opposite/hypotenuse

sin(55) = x/25

25*sin(55) = x

x = 25*sin(55)

x = 20.4788011072248

x = 20.5

Each pole is roughly 20.5 feet tall.

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Problem 6

The exact value is the fraction sqrt(3)/2 which can be written as \frac{\sqrt{3}}{2}

"sqrt" is shorhand for "square root"

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Answer 2

Problem 1

cos(angle) = adjacent/hypotenuse
cos(18) = x/25
25*cos(18) = x
x = 25*cos(18)
x = 23.7764129073789
x = 23.78

Answer: x = 23.78

=====================================================

Problem 2

theta = greek letter that looks like a "zero" or the letter O with a horizontal line through it

With the reference angle of theta, the opposite side is 3. This side is the leg furthest from the angle theta. In contrast, the side 4 is the closest leg to theta. The longest side 5 is the hypotenuse. The hypotenuse is always the lonest side

sin(theta) = opposite/hypotenuse
sin(theta) = 3/5

cos(theta) = adjacent/hypotenuse
cos(theta) = 4/5

tan(theta) = opposite/adjacent
tan(theta) = 3/4

---------------------------

In summary:
sin(theta) = 3/5
cos(theta) = 4/5
tan(theta) = 3/4

=====================================================

Problem 3

cos(angle) = adjacent/hypotenuse
cos(x) = 15/29
x = arccos(15/29) .... inverse cosine or cos^(-1) also works
x = 58.852610078491
x = 58.85

Answer: x = 58.85

=====================================================

Problem 4

The rule we'll use is
sin(x) = cos(90-x)

sin(theta) = cos(90-theta)
cos(90-theta) = cos(58)
90 - theta = 58
90 - theta+theta = 58+theta
90 = 58+theta
90-58 = 58+theta-58
32 = theta
theta = 32

Answer: 32

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Problem 5

x = height of pole

the vertical side is the side we want, which is the opposite side compared to the angle 55 degrees

sin(angle) = opposite/hypotenuse
sin(55) = x/25
25*sin(55) = x
x = 25*sin(55)
x = 20.4788011072248
x = 20.5

Each pole is roughly 20.5 feet tall.

=====================================================

Problem 6

The exact value is the fraction sqrt(3)/2 which can be written as $\frac{\sqrt{3}}{2}$

"sqrt" is shorhand for "square root"