Thank you, that was very good. Though, for part c), since you'd already figured out that alpha = 80 degrees, you could have bypassed using the cosine rule, and recognised that the internal angles of a triangle = 180 degrees. From there, we have 180 - 80 - 70 = 30 degrees. That said, I really appreciated your instruction, I was never taught true bearings and found it a bit confusing, so thank you for clearing that up for me.
I kinda getting it but u are skipping out on some steps and your voice going low when explaining at some point like u don't want to talk up..... explain a lil better. Not getting it
Hi Alice To find x with the 70 degrees we use the cosine rule for non right angled triangles, it looks a bit like pythagoras but has a bit on the end. hope this helps bob
Hi, I will makea more fully explained video, if you could, if you have time, let me know the bits that you feel need more explanations and I will incorporate them in the new video. Many thanks Bob
mathswithbob Hi, thanks for the response. I just meant that when you were able to say ‘we can tell this angle is x degrees’, if you could explain how you can tell.
Man last angle could be so easily calculated We have 80 in the triangle plus 50 so just add them to get 130 (second quadrant and the tiny angle we need) the total angle then u just could have subtracted the second quadrant(90) from the total to get the frickin 40
Wouldn't it have been easier to use the knowledge that the internal angles of a triangle = 180? 180 - 80 - 70 = 30 Then 90 - (30+20) = 40 But hey, let's not fight on the internet.
For the last angle, drawing a right angle triangle and using the side length x for the hypotenuse might have been a quicker way of working out the angle. With this, you could say that sinB is equal to opposite/x. Since we know the value of x, we do xsinB, which makes it equal to opposite. Once we've found the opposite side length, we can do the inverse of sin with the values opposite/x, knowing both the values. This would be an alternative way of finding the angle, which also teaches people watching your video to be able to use that same method in another problem. Of course, the quickest way of doing it is 180 - a - 50 - 20, since all triangles add up to 180. While the way you solved it drilled it into everyone's heads through repetition, there were more useful ways to solve the last question.
Hi Bob...can you explain how you can be using Pythagorus on a non-right-angle triangle? your figures clearly state the angle opposite to AC is 70 deg not 90...or am I missing something?
Hi Shankar The first 50 comes from 140-90 at A, ( or how far past 90 do we need to go to get the bearing 140T ) Then the 50 at B is the alternate angle with parallel EW lines. Bob
I don’t want to be mean but I think you should take down this video and re-do it step by step by more steps identifying and stating angle rules for obtaining some of your values. It is so confusing where you get values from and why? If it wasn’t for the comments I would be absolutely lost. 🤷♀️
Hey, you may try this ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-XlkT1iZO2HI.html GCSE Maths True bearings | How to find bearing | Finding Bearings | Best Examples It might help U better
Hi Shamel, With true bearing all the north directions are parallel, ( at each port draw new Norths ) this creates alternate angles ( they are equal ) with the bearing lines. There is a lot of non right angled trig involved ( cosine and sine rules ). If you have the time could you let me know which bits could be explained better to be more helpful. Thanks Bob
