1. anonymous

ok

2. anonymous

Sure

3. anonymous

#208 only. :) thankyou.. ^_^

4. anonymous

Can you please type the question so it is easier to see

5. anonymous

oww.. okay.. sure. :) wait a minute..

6. anonymous

Can you see it clear?

7. anonymous

Yep thanks let me find the answer

8. anonymous

thankyou.. :) i'll try to answer it again. ^_^

9. anonymous

Hey there. We will have to use Pythagoras here, part of Mechanics exercises.

10. anonymous

ahh.. but how to illustrate the several forces?

11. anonymous

Relax, I will show you.

12. anonymous

For a, you will have diagrammatically, the following representative situation |dw:1434984303367:dw|

13. anonymous

Finding the hypotenuse will be the resultant force: $DeltaP=\sqrt{(-200)^{2}+100^{2}}=223.7=224N$ For the angle, we can use Sine function, giving us$\sin \beta=\frac{ 200 }{ 224 }$ $\beta=\sin^{-1} \frac{ -200 }{ 224 }=-63.2$ Given that you do not want negative angles, we use the property 90-B=90-63.2=26.7Degrees. Can you use the same methodology for the other exercises?

14. anonymous

What do you mean?

15. anonymous

|dw:1434985341630:dw|