# TrigonometryBasic trig, solving side of triangle

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#### daigo

##### Member
A ship is headed east @ 15 MPH.
There is a tide North-10 degrees-East @ 5 MPH.
Find the course and speed of the ship.
Here is what I got so far:

But the solution 15.7875 MPH is incorrect so undoubtedly my angle would come out incorrect too.

Where have I gone wrong?

#### Sudharaka

##### Well-known member
MHB Math Helper
Here is what I got so far:

But the solution 15.7875 MPH is incorrect so undoubtedly my angle would come out incorrect too.

Where have I gone wrong?
Hi daigo,

You have only considered the vertical component of the velocity of the tide. What about the horizontal component of the tidal velocity? It should be added to the velocity of the ship.

Kind Regards,
Sudharaka.

#### biffboy

##### New member
Let i be unit vector in direction east and j be uit vector in direction north
Tide is vector 5sin10i+5cos10j
Boat vector is 15i
resultant is (15+5sin10)i+5cos10j so you want the magnitude and direction of this vector
Note For vector ai+bj magnitude is root(a^2+b^2) and angle to j direction is given by tanx=a/b

#### daigo

##### Member
Is there a way to solve this without velocity or vectors? (If we get rid of all the units and just consider the numbers)

I just used the Pythagorean theorem for side '15' and side '5cos(10)'

15 must be correct because it's a given, so 5cos(10) must not be the correct length of the other side. I can't see why that is, though

#### Jameson

Staff member
You are asked to find the course, which in this case means direction. You can find the speed by using the Pythagorean Theorem but you still need to find the direction first. Normally the y-component of a vector is found using sine, not cosine as you did and I don't see why that rule would change so try using sine. Once you have all of your components, add the x-components and y-components together to get the resulting velocity.

Ship velocity vector: <15, 0>
Tide velocity vector: <x, y>

Combined velocity: <15+x, 0+y>

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#### daigo

##### Member
Jameson, I am reading every single post and trying to consolidate the information in a way I can understand so I can apply it, but I am just not understanding. I wish I wasn't so slow at maths

The side I labeled 'x' as I understand it, is the final direction and speed in which I believe the ship to be travelling

I will label the appropriate vectors in a new drawing, please hold

#### Jameson

Staff member
Jameson, I am reading every single post and trying to consolidate the information in a way I can understand so I can apply it, but I am just not understanding. I wish I wasn't so slow at maths
Hi daigo!

I didn't mean that to come off as an insult, I just thought you really didn't see my posts since you didn't ask any questions or comment on them. I do apologize if it seemed otherwise. I really like how you're trying to figure this out on your own, honestly! It shows that you are trying to really understand the concepts instead of memorizing them. That sounds to me like the opposite of slow at maths.

Jameson

EDIT: The way you've labeled the diagram is the way I interpreted how you were setting it up. I still think there is a problem with the final course part. You drew the tidal vector correctly, so now attach the ship's vector to that such that the ship vector starts where the tidal vector ends. Then draw the resulting vector from the origin to where that ends.

Does that make sense?

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#### daigo

##### Member
EDIT: The way you've labeled the diagram is the way I interpreted how you were setting it up. I still think there is a problem with the final course part. You drew the tidal vector correctly, so now attach the ship's vector to that such that the ship vector starts where the tidal vector ends. Then draw the resulting vector from the origin to where that ends.

Does that make sense?
Ah, I guess that this is similar to the "head to tail" method or whatever it's called. That method did not make sense to me intuitively. So you are saying:

Why this does not make sense to me is because this is how I am thinking about it:

A boat is trying to move eastbound, and a tide is pushing the boat northbound, so it travels in a somewhat diagonal path. In my mind, that would mean the the boat is being pushed faster than its initial velocity of 15 MPH, so I know the speed will be higher than that. I think I am confusing the velocities with distance a bit, but I guess I can just accept the fact that you just add velocities even though it doesn't make sense to me that you add them for some reason. Because if you add the speed together, wouldn't it just be 20 MPH? But I guess the angle slows the speed down a bit because the boat and the tide are not headed in the exact same direction.

#### daigo

##### Member
I guess this is another way I am thinking of it:

#### Jameson

Staff member
That's a decent approximation for the general direction but it doesn't give you any of the answers you need. If that's just for your understanding then yep, that's the direction it will be roughly heading.

I think I have explained all that I can without starting to repeat myself so I suggest learning how to use the vector component method of addition to get the solution to this problem then continue drawing diagrams for problems until you feel more comfortable with it. The more you do everything correctly, the more your intuition will adjust and it won't seem weird anymore.

EDIT: Just to wrap up this thread and address the original problem, the resultant vector is $$\displaystyle <15+5\cos(80), 5\sin(80)>$$ as biffboy said earlier in the thread. Note that this is in terms of degrees, not radians for when you calculate the magnitude of the resultant velocity, or the speed.

Also as biffboy mentioned to get the direction you must use an inverse trig function, the easiest being $$\displaystyle \tan^{-1}$$

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#### daigo

##### Member
For example:

Three people are pulling on ropes tied to a rock, each person at a force of 75 lbs., 100 lbs., and 125 lbs. at angles of 30 degrees, 45 degrees, and 120 degrees respectively with the positive x-axis.
To find the vector and direction I can just do:

v_1 = 75<cos30,sin30>
v_2 = 100<cos45,sin45>
v_3 = 125<cos120,sin120>

v_1 = <65,38>
v_2 = <71,71>
v_3 = <-63,108>

||v|| = sqrt(73^2+216^2)

229 lbs. @ arctan(216/73) = 71 degrees

I can estimate that this solution is correct, but I have no idea why it came to this and how. I just know that mathematically, it all works out in the end. Maybe it's just something I should just accept and not concern myself about the abstraction/analytical conceptualization?

#### Jameson

Staff member
Again if you think about this in terms of the horizontal force and vertical force it's easier to see how they affect each other because when adding components anything with the same sign will increase the force, velocity, etc. in that direction and adding things with different signs will weaken or slow down this component.

You can easily understand how forces add together if there are only two directions, left/right...up/down...plus/minus or whatever. In all of these cases you can easily and intuitively get why 2N up and 2N down cancel out. Everything is about sign. So given that intuition if you can understand the math behind rewriting a vector in terms of its components then you should trust the whole method more because each component deals with only two directional choices which are opposites.

If you mean being able to look at a list of vectors with different directions/magnitudes as well as involving all the quadrants and instantly be able to picture the result of adding them together then I would think that comes with practice and can be used to double check the answer makes sense, but that's not something you are expected to know immediately so don't feel like it's just you.

I think it's officially time for me to step back from this discussion and hopefully allow other, more knowledgeable members to explain this perhaps in way that suits you more. I hope some of these ramblings have helped!

#### Sudharaka

##### Well-known member
MHB Math Helper
A ship is headed east @ 15 MPH. There is a tide North-10 degrees-East @ 5 MPH. Find the course and speed of the ship.
Hi daigo,

I hope Jameson's countless number of posts would have given you a good insight about this problem. In case if you still have some doubts left, let me explain the way I would think about this.

This is the initial figure that you can draw from the given details.

Now let us consider the components of each velocity in the horizontal(East) and vertical(North) directions.

1) The ship has a velocity component of 15 mph in the horizontal direction and a velocity component of 0 mph in the vertical direction.

2) The tide has a velocity component of $$5\sin(10^0)$$ mph in the horizontal direction and a velocity component of $$5\cos(10^0)$$ mph in the vertical direction.

To find the resultant velocity of the ship, we have to add the two velocities: velocity of the ship and the velocity of the tide. We have split each velocity into components so that we can add the corresponding components to get the horizontal and vertical components of the resultant velocity.

Therefore,

Horizontal component of the resultant velocity(in mph) = $$15+5\sin(10^0)$$

Vertical component of the resultant velocity(in mph) = $$0 + 5\cos(10^0)$$ = $$5\cos(10^0)$$

Now we know the two components of the resultant velocity. What is left for us to do is to find the magnitude and the direction of the resultant velocity. We shall use the parallelogram law of vector addition(see this and this) to find the resultant vector.

By the Pythagorean theorem, we can find the magnitude of the resulting vector.

Magnitude of the velocity of the ship = $$\sqrt{(15+5\sin(10^0))^2+(5\cos(10^0))^2}\approx 16.61\mbox{ mph}$$

Also, $$\displaystyle\theta = \tan^{-1}\left(\frac{15+5\sin(10^0)}{5\cos(10^0)}\right)=\tan^{-1}\left(\frac{3+\sin(10^0)}{\cos(10^0)}\right) \approx 72.76^0$$

Therefore the ship moves $$72.76^0$$ North of East with a speed of 16.61 mph.

Kind Regards,
Sudharaka.