1)a line segment (not a side) connecting two vertices of a polygon is called a _______.
2 answers:
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Answer:
Step-by-step explanation:
So we have the equation:
To solve, subtract 5 from both sides. This is the subtraction property of equality:
The left side cancels:
Subtract on the right.
So, our answer is -7.
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Answer:
6. (A, B, C) ≈ (112.4°, 29.5°, 38.0°)
7. (a, b, C) ≈ (180.5, 238.5, 145°)
Step-by-step explanation:
My "work" is to make use of a triangle solver calculator. The results are attached. Triangle solvers are available for phone or tablet and on web sites. Many graphing calculators have triangle solvers built in.
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We suppose you're to make use of the Law of Sines and the Law of Cosines, as applicable.
6. When 3 sides are given, the Law of Cosines can be used to find the angles. For example, angle A can be found from ...
A = arccos((b² +c² -a²)/(2bc))
A = arccos((8² +10² -15²)/(2·8·10)) = arccos(-61/160) = 112.4°
The other angles can be found by permuting the variables appropriately.
B = arccos((225 +100 -64)/(2·15·10) = arccos(261/300) ≈ 29.5°
The third angle can be found as the supplement to the other two.
C = 180° -112.411° -29.541° = 38.048° ≈ 38.0°
The angles (A, B, C) are about (112.4°, 29.5°, 38.0°).
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7. When insufficient information is given for the Law of Cosines, the Law of Sines can be useful. It tells us side lengths are proportional to the sine of the opposite angle. With two angles, we can find the third, and with any side length, we can then find the other side lengths.
C = 180° -A -B = 145°
a = c(sin(A)/sin(C)) = 400·sin(15°)/sin(145°) ≈ 180.49
b = c(sin(B)/sin(C)) = 400·sin(20°)/sin(145°) ≈ 238.52
The measures (a, b, C) are about (180.5, 238.5, 145°).
