Ello, estoy aquí para ayudar, pero hablo español con optimismo, comprenderá lo que estoy diciendo. Intentaré ayudarlo. Espero que tengas un buen dia <3
Step-by-step answer:
Referring to the attached diagram, the resultant of two forces each with magnitude F and inclined to each other at 2a equals
Ra = 2Fcos(a) ..............................(1)
Similarly, the resultant of two forces each with magnitude F and inclined to each other at 2b equals
Rb = 2Fcos(b)..............................(2)
We are given that
Ra = 2Rb ....................................(3)
Substitute (1) & (2) in (3) gives
2Fcos(a) = 2(2Fcos(b))
Expand
2Fcos(a) = 4Fcos(b)
Simplify
cos(a) = 2 cos(b) QED
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Have a nice day!
Triangles JFE and JLK are similar provided that line segments FE and LK are parallel. This means that corresponding parts of these triangles occur proportionally to one another. In particular,

From this relation we get

Now we can solve for
:




Answer:
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Step-by-step explanation:
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