Angle needed for object A to intercept with object B

Question

Object B is 15 degrees East of North at a distance of 20km/h. Object B is moving at an average speed of 30km/h in the direction 40 degrees East of North. If object A is capable of moving at 100km/h, at what angle does it need to move to intercept object B?

I've tried calculating it as vectors, but got stuck. In the end I just tried using good old geomatry, where $$ \frac{\sin A}{a} = \frac{\sin B}{b} $$ So I called the angle I'm searching for A, and the angle from the line between A and B, and the direction of object B, as angle B. 'a' and 'b' are the average speeds of object a and b. $$ \frac{\sin A}{30 \times t} = \frac{\sin(155°)}{100\times t} $$ $$ A = arcsin\left(\frac{30\times t\times \sin(155°)}{100\times t} \right) = 7.284° $$

So the final angle should be $15°+7.284°=\mathbf{22.284°}$

Is this correct? Isn't there a better way of doing this, like using vector products?

PS. I've found a very similar question here, but didn't quite understand how i.e. $\alpha + \beta = 40° $, because my calculations show $ =35° $. And trying to read through comments and answers, confuses me more due to updates being made inconsistent to comments about the updates done. And due to me not having enough reputation, I'm not able to ask counter-questions.

BONUS Question: I have been searching high and low on how to make a simple graph of my problem. Like this here question, with a graph of the problem hosted at stack.imgur.com which I guess was generated by LaTeX on stackexchange.com, right?

EDIT: Added obligatory sketch!

Answer 1

Simple way to solve: Use distances travelled. So B has already travelled 20km at 15° East of North. That is $20 \times \sin(15°)$km northwards and $20 \times \cos(15°)$ eastwards.

Now consider A intercepts B at a time $t$. The equations for the distances travelled by B are:

$Y_B = 20 \times \sin(15°) + 30 \times t \times \sin(40°)$ in the Y direction and
$X_B = 20 \times \cos(15°) + 30 \times t \times \cos(40°)$ in the X direction.

For A, the distances travelled are:
$Y_A = 100 \times t \times \sin(\alpha)$ in the Y direction and
$X_A = 100 \times t \times \cos(\alpha)$ in the X direction.

For the two to intercept, $X_A = X_B$ and $Y_A = Y_B$

You now have 2 equations, 2 variables, and a whole lot of fun solving them simultaneously.

Answer 2

There is a technique based on vectors that solves the problem, and lets you know if there is no solution, one solution or two solutions.

(All my angles are measured counter-clockwise from the positive X-axis)

Assume the pursuer travels at 100 km/hr at an angle $\beta$. We're going to try to divide this velocity into two non-perpendicular components

One matches the velocity of the target exactly. So in the problem above, assume that one part of A's velocity is $30\angle 50°$. It wouldn't matter if A's total velocity were less than 30.

So, we have (in theory) completely cancelled out B's attempt to escape. Now, how does A reach B? A must travel at some velocity (say X) along the original direction from A to B. (Since we've cancelled out any of B's movement) So, in this problem, the other part of A's velocity is $X \angle 75°$

So, finally, we're reduced to solving:$$30\angle 50°+X \angle 75°=100\angle\beta$$One way is to expand each vector into X and Y components and set up two equations. If you eliminate $\beta$ you get a quadratic in X. If there are any real positive values of X, you can then substitute and solve for $\beta$

ADDITIONAL DETAILS:

Expanding the equation above for components in the X and Y direction: $$30\cos(50)+X\cos(75)=100\cos(\beta)$$ $$30\sin(50)+X\sin(75)=100\sin(\beta)$$The only unknowns are the size of X (or if it even exists!) and the angle $\beta$ If you square both sides of both equations and then add the two LHS and two RHS, some trig manipulations will eliminate $\beta$ and lead to the quadratic in X.