Calculating the Angle between Ship and Rock

[evinda]

Hello! (Wave)

A ship that is at the position $(1,0)$ of a chart ( with the north at the positive direction $y$) "sees" a rock at the position $(2,4)$. Which is the vector that connects the ship with the rock? What angle $\theta$ does this vector form with the direction of the north?

I have thought to draw the x and y-axis:

View attachment 5319

The vector that connects the ship with the rock is $(1,0)+ t (1,4), t \in \mathbb{R}$, isn't it?We consider the triangle:

View attachment 5320Do we have to prove that it is a right triangle in order to use for example that $\sin {\phi}= \frac{\text{opposite}}{\text{hypotenuse}}$ ?

 

[MarkFL]

I would say that in a Cartesian coordinate system, we may freely use the fact that the lines $y=k_y$ and $x=k_x$ are perpendicular, just as we do when using the distance formula for example.

 

[evinda]

I would say that in a Cartesian coordinate system, we may freely use the fact that the lines $y=k_y$ and $x=k_x$ are perpendicular, just as we do when using the distance formula for example.

View attachment 5321

So do we just say that since we have a right triangle, it holds that $\sin{\phi}=\frac{(2,4)-(2,0)}{(2,4)-(1,0)}=\frac{(0,4)}{(1,4)}$ ?

What is meant with "the direction of the north" ?

 

[MarkFL]

Since we know without using the Pythagorean theorem the lengths the of legs, I would use the tangent function instead. And they are asking for the angle the vector makes with the northern direction, or the $y$-axis...let's call this angle $\beta$, and so we could use:

\(\displaystyle \beta=\frac{\pi}{2}-\arctan\left(4\right)\)

 

[evinda]

Since we know without using the Pythagorean theorem the lengths the of legs, I would use the tangent function instead. And they are asking for the angle the vector makes with the northern direction, or the $y$-axis...let's call this angle $\beta$, and so we could use:

\(\displaystyle \beta=\frac{\pi}{2}-\arctan\left(4\right)\)

Could you explain it further to me? (Sweating)

 

[MarkFL]

Could you explain it further to me? (Sweating)

Draw the vertical line $x=1$. Now, can you see that the angle subtended by this vertical line and the line segment from $(1,0)$ to $(2,4)$, which I have called $\beta$ is complementary to the angle you have labelled $\phi$?

We could get the angle we want more directly though...please consider the following diagram:

View attachment 5322

And so we see:

\(\displaystyle \beta=\arctan\left(\frac{1}{4}\right)\)

And this is equivalent to the value I posted before (do you see why?) :)

 

[evinda]

\beta=\arctan\left(\frac{1}{4}\right)[/MATH]

And this is equivalent to the value I posted before (do you see why?) :)

So we can pick this right triangle instead of the one that I drawed, right?

It holds that $\tan \beta=\frac{\text{opposite}}{\text{adjacent}}=\frac{1}{4} \Rightarrow \beta=arc \tan{\frac{1}{4}}$, right?

Why is $arc \tan{\frac{1}{4}}$ equal to $\frac{\pi}{2}-arc \tan{4}$ ?

 

[evinda]

I am looking again at the question [m] Which is the vector that connects the ship with the rock? [/m].Isn't the vector equal to (2,4)-(1,0)=(1,4) ?

 

[MarkFL]

So we can pick this right triangle instead of the one that I drawed, right?

Yes, and in fact it is congruent to the one you drew.

It holds that $\tan \beta=\frac{\text{opposite}}{\text{adjacent}}=\frac{1}{4} \Rightarrow \beta=arc \tan{\frac{1}{4}}$, right?

Correct. :D

Why is $arc \tan{\frac{1}{4}}$ equal to $\frac{\pi}{2}-arc \tan{4}$ ?

Consider the identities:

\(\displaystyle \arctan\left(\frac{a}{b}\right)=\arccot\left(\frac{b}{a}\right)\tag{1}\)

This comes from that fact that tangent and cotangent are multiplicative inverses of each other.

and

\(\displaystyle \arctan(a)+\arccot(a)=\frac{\pi}{2}\tag{2}\)

This comes directly from the fact that tangent and cotangent are complementary functions.

And so, let's begin with:

\(\displaystyle \theta=\arctan(a)\)

Applying (2), we may write:

\(\displaystyle \theta=\frac{\pi}{2}-\arccot(a)\)

And then applying (1), we obtain:

\(\displaystyle \theta=\frac{\pi}{2}-\arctan\left(\frac{1}{a}\right)\)

 

[MarkFL]

I am looking again at the question [m] Which is the vector that connects the ship with the rock? [/m].Isn't the vector equal to (2,4)-(1,0)=(1,4) ?

Yes, that's correct. (Yes)

 

[evinda]

View attachment 5323Why do we consider that the line that passes through the points $(1,4)$ and $(2,4)$ is $1$ and not the vector $(1,0)$ ?

 

[evinda]

Or do we mean the length of the opposide and the length of the adjacent vector?

Also could you explain to me why $\beta$ is the angle that the vector $(1,4)$ forms with the direction of the north?

 

[MarkFL]

Why do we consider that the line that passes through the points $(1,4)$ and $(2,4)$ is $1$ and not the vector $(1,0)$ ?

The length of that side of the triangle is equal to the magnitude of $\langle 1,0 \rangle$:

\(\displaystyle \left|\langle 1,0 \rangle\right|=\sqrt{1^2+0^2}=1\)

I actually simply took the magnitude of the difference in $x$-coordinates.

 

[evinda]

The length of that side of the triangle is equal to the magnitude of $\langle 1,0 \rangle$:

\(\displaystyle \left|\langle 1,0 \rangle\right|=\sqrt{1^2+0^2}=1\)

I actually simply took the magnitude of the difference in $x$-coordinates.

Since it's the line $(1,0)$ we can just say that its length is 1, can't we?

 

[MarkFL]

Or do we mean the length of the opposide and the length of the adjacent vector?

Yes, which is what I was saying in my previous post (that I posted after your question). :)

Also could you explain to me why $\beta$ is the angle that the vector $(1,4)$ forms with the direction of the north?

The direction of north is "up", or the same as the positive $y$-axis, and so it has the same direction as the vector:

\(\displaystyle \langle 1,4 \rangle - \langle 1,0 \rangle=\langle 0,4 \rangle\)

 

[evinda]

Yes, which is what I was saying in my previous post (that I posted after your question). :)

Ok... (Smile)

The direction of north is "up", or the same as the positive $y$-axis, and so it has the same direction as the vector:

\(\displaystyle \langle 1,4 \rangle - \langle 1,0 \rangle=\langle 0,4 \rangle\)

What does $\beta$ represent? Why is it the angle that $(1,4)$ forms with the y-axis?
How would we find for example the angle that $(1,4)$ forms with the x-axis?

 

[MarkFL]

...What does $\beta$ represent? Why is it the angle that $(1,4)$ forms with the y-axis?
How would we find for example the angle that $(1,4)$ forms with the x-axis?

Please refer to the diagram I posted in post #6...we are defining the angle $\beta$ to be that which is subtended by the vectors $\langle 0,4 \rangle$ and $\langle 1,4 \rangle$...wouldn't you agree this is the angle for which we are asked? After all, doesn't the vector $\langle 0,4 \rangle$ point due north?

 

[evinda]

Please refer to the diagram I posted in post #6...we are defining the angle $\beta$ to be that which is subtended by the vectors $\langle 0,4 \rangle$ and $\langle 1,4 \rangle$...wouldn't you agree this is the angle for which we are asked? After all, doesn't the vector $\langle 0,4 \rangle$ point due north?

Ok. And no matter which other right triangle we would draw we would get the same result, right?

Is the use of a right angle the only way to find the desired angle?

 

[MarkFL]

Ok. And no matter which other right triangle we would draw we would get the same result, right?

Is the use of a right angle the only way to find the desired angle?

Yes, we are not locked into any particular triangle, but as long as properly done, you will get the same result.

We could use the Law of Cosines, or vector algebra, or analytic (coordinate geometry), but they are all really just variants of this same method. :)