What additional information can be used to prove ΔEFG ≅ ΔYXW by SSS?
2 answers:
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Answer:
138.06 lb N86.1°E
Step-by-step explanation:
There are several ways you can work this. One of the most straightforward is to resolve each vector into its north and east components, add those, and then covert the result back to magnitude and direction.
A diagram can help immensely.
We note that S78°E is the same as E12°S. Since we're used to seeing the coordinate system with the +x axis aligned with east, it can be convenient to think of the first force as 100 lb at -12°.
The angle of the second force is N53°E, which can be expressed as E37°N. Then in x-y coordinates, this force is 50 lb at +37°.
The components of the sum are the sum of the components:
R = F1 +F2 = (100cos(-12°) +50cos(37°), 100sin(-12°) +50sin(37°))
= (137.747, 9.300)
The the magnitude of the resultant is computed using the Pythagorean theorem:
||R|| = √(137.747² +9.300²) ≈ 138.06
and the angle is computed using the arctangent function. Here, our diagram tells us the angle is in the first quadrant, so is positive (relative to +x, or East).
∠R = arctan(9.300/137.747) ≈ 3.9°
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We want to express the answer in terms similar to the way the given forces are expressed, so we want the angle relative to north. The resultant is then described by ...
R = 138.06 lb at N86.1°E
11/12 of 7 is simply their product.