Answer:
therefore the left work of worker will be 6/7 part of work
1. What is the value of x in the triangle?
2 answers:
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
=====================================================
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
"sqrt" is shorhand for "square root"
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
=====================================================
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
"sqrt" is shorhand for "square root"
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