# Finding resultant and direction of resultant

In this question-

A motorboat is racing towards the north at 25km/h and the water current in that region is 10km/hr in the direction of 60 degrees east of the south. Find the resultant velocity of the boat.

The first part is quite easy and we get $$21.8$$ approx as the magnitude of the resultant.

My doubt is in the 2nd part of the question. How do we calculate the direction using the normal method. One way is to use the sine formula and say that

$$\frac{R}{\sin(a)}=\frac{\text{velocity of current}}{\sin(\alpha)}$$

where $$a$$ is the angle between the vectors and $$\alpha$$ is the angle of resultant $$R$$ with the north direction. this gives us the angle with the north that is $$23.4^\circ$$ which is correct but how does one use the normal $$\tan(\alpha)$$ method which works in questions where $$\alpha$$ is less than $$90$$ or equal to $$90$$.

By the $$\tan(\alpha)$$ method I am referring to this formula-

$$\tan(\alpha)=\frac{A\sin(a)}{b+A\cos(a)}$$

One of the books says that it is

$$\tan(\alpha)=\frac{10\sin120^\circ}{25+10\cos120^\circ}$$

Which gives the correct answer also but I can't understand how we get this. After making a parallelogram, if the angle between vectors is less than $$90$$ or $$90$$, we just extend one of the sides and get 2 right-angle triangles. In this case, we get only one. Please explain how do we do this.

• How can you get the length of the vector without knowing the direction? And does "towards north" mean that it is heading (i.e. pointing) North, or that its velocity w.r.t. the earth is North (i.e. that it is the resultant that is pointing North). "At 25 km/h" - is that with respect to the water, or the land? It is decidedly unclear. Perhaps if you draw a diagram (ALWAYS draw a diagram) we can see what you mean. And when you have done so, the answer will become quite obvious, I think... Just compute the resultant velocity to the East and North separately. Sep 12, 2014 at 19:53 