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!

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!