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Can you please explain me how have we resolved v (rain,road) for the second diagram as
v(rain,road)sin(30°+a)=12km/h cos30° in detail ?

Can you please explain me how have we resolved v (rain,road) for the second diagram as

v(rain,road)sin(30°+a)=12km/h cos30° in detail ?

### Asked by Varsneya Srinivas | 12th May, 2017, 04:31: PM

Expert Answer:

###
- Taking the components along OA indicates that OA has to be taken as a base reference to resolve the forces around it which are mentioned in equation (i).
- When OA is taken as a base reference, it is found that by the geometry of the velocity v man,road=2kmph, OA happens to be the horizontal component of this velocity.
- Therefore, OA=12kmph.Cos30.

- Let the quadrilateral in the above diagram be parallelogram OPQR.
- In triangle OQA, angle Q=(30+a) as angle Q and angle (30+a) are the alternate angles formed between two parallel sides OP and RQ.
- Now by applying trigonometry in triangle OQA

- This is how we get equation (iii) of the solution mentioned in your query.

- This is how we get equation (iii) of the solution mentioned in your query.

### Answered by Abhijeet Mishra | 14th May, 2017, 12:44: PM

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