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Based on the construction, we can logically deduce that: A. yes; the compass was kept at the same width to create the arcs for points C and D.
<h3>What is a line segment?</h3>A line segment can be defined as the part of a line in a geometric figure such as a triangle, circle, quadrilateral, etc., that is bounded by two (2) distinct points and it typically has a fixed length.
In Geometry, a line segment can be measured by using the following measuring instruments:
- A scale (ruler)
- A divider
- A compass
In Geometry, an arc can be defined as a trajectory that is generally formed when the distance from a given point has a fixed numerical value.
Based on the construction with arcs created above and below the line segment from points A, we can infer and logically deduce that it is true that the compass was kept at the same width to create the arcs for points C and D.
In conclusion, yes, the construction demonstrated how to bisect a line segment correctly by hand.
Read more on arcs here: brainly.com/question/11126174
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Complete Question:
The construction has a given segment AB. Arcs have been created above and below the segment from points A that are equidistant from point A. The compass was kept at the same distance, placed on point B, and two additional arcs were created above and below the segment that intersect with the first arcs created. The intersection of the arcs above the segment created point C. The intersection of the arcs below the segment created point D. A line was drawn from point C to D through the segment.
Does the construction demonstrate how to bisect a segment correctly by hand?
A. Yes; the compass was kept at the same width to create the arcs for points C and D.
B. Yes; a straightedge was used to create segment CD.
C. No; the compass was not kept at the same width to create the arcs for points C and D.
D. No; a straightedge was used to create segment CD.
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Answer:
31.243 units
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you of the relationships between sides and angles in a right triangle. Using the attached figure, it is convenient to find the length of BE as an intermediate step in the solution.
Sin = Opposite/Hypotenuse
sin(30°) = BE/100
BE = 100·sin(30°)
Then ...
Tan = Opposite/Adjacent
tan(58°) = BE/x
x = BE/tan(58°) = 100·sin(30°)/tan(58°)
x ≈ 31.243 . . . . units
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<em>Comment on the figure</em>
The intermediate problem in creating the figure was to locate point D. That was accomplished by locating point C on a line at an angle of 58° CCW from the horizontal, using point B as a center. Then D is the intersection of BC with the x-axis. BE is drawn perpendicular to the x-axis.
