Two perpendicular forces have a resultant of 100N, if one of the forces is at an angle 60 with the resultant force find the magnitude. Please am so confused that I even think the question is wrong

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The magnitude of that one force?

Make 100 N the length of that resultant force, and you can rearrange the forces in that way:


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And from that triangle and the known angle, you can use trigonometry to find the magnitudes of all the other forces concerned.

Can you do that?

its just kind of hard for me. I know that R^2=x^2+y^2. I think R which is the resultant here is given to be 100N, so I think am to find the value of the two other forces at right angle of which one of the forces is at an angle 60 with the resultant R. that's where an numb

There's trigonometry that you need too. If you have a scientific calculator, you'll see the options sin, cos, tan. Those are what you need. Otherwise, if you did them in your math class, you could also know the exact values.

http://p1cture.it/images/e5gd0ae3f3e91gcf1a35.png

Okay, for the ratios that you have:

\sin(60^o) = \frac{y}{100}

\cos(60^o) = \frac{x}{100}

You can put sin(60) in your calculator, and you should get 0.8660254... (the exact value is \frac{\sqrt{3}}{2}) and if you type in cos(60), you should get 0.5. From that, you can get any of the two forces which contribute to the resultant force :)

PS. This works ONLY for right angled triangles.

Thanks you just did the majic

You're welcome! :)

Thanks for having left a thumbs up :)