To clarify this doubt, first read the question carefully what is given and second understand the difference between bearings and included angles. Bearings represent the direction of a line with respect to North so line AB is 60 degrees from the North.... whereas the angles that you have mentioned, are all included angles i.e. angle between two consecutive lines and it has nothing to do with actual direction. So, these angles (A, B, C & D) have nothing to do with quadrants. Now, try to make the figure by yourself and see the result (FYI: Here North is Magnetic North)
Sir, regarding station C, the angle looks like it is actually 60 degrees 22'. But if we check it from North direction it it NOT lying in first quadrant. Can you please explain?
No...It seems most of the students are not understanding the question properly. First, understand the difference between a bearing and an included angle. Bearing is the angle taken from the North direction whereas an included angle is simply an angle between two consecutive lines inside of a closed polygon. So, here AB 60 degrees, is the angle from the North whereas 60 degrees and 22' is just an angle between two consecutive lines that has nothing to do with North direction. So, don't associate the word "Quadrant" with included angles rather use these included angles to determine in which Quadrant a particular line lies.
Sir how to distribute error in bearing if there is a error in total included angle. Can we distribute the error in first bearing or only on the other three bearings.
If sum of all the included angles is not equal to (2n-4)*90 where n is number of sides, then find the error and divide that error with number of included angles. After dividing, distribute the error in each included angle. Remember, close the traverse first and then fix the bearings.
sir how to calculate it as azimuth? its going to an open skitch or closed? imean for point B to C will be draw 3rd point from 2nd point to 3rd or from first to another 2nd point down? sorry for bad english i hope u answer soon,many students watch you from kurdistan. thanks
sir,say if i started clockwise first comes bearing BC which is equal to bearing AB(i.e, 60°) +included angle B(i.e,90°8')=150°8' but we got bearing BC=149°52' doesn't seem right.......plzz tell me how to calculate bearings if we proceed clockwise
#Shalom Christie ...sorry for the late reply. You wrote "bearing BC which is equal to bearing AB(i.e, 60°) +included angle B(i.e,90°8')=150°8'" . I'm afraid this is an incorrect approach. I strongly suggest you go through the entire compass surveying playlist and all your doubts will be cleared.
You can start from either direction ! Your bearings will be same...whatever direction you choose, make sure that at the end (as a check) the bearing of line AB must be 60 degrees.
@@CivilModules aren’t whole circle bearings just Azimuths? I’m just confused because my professor said if our bearings are larger than 90 then it’s wrong.
No..no need to be confused. Yes.. Technically Bearing refers to the readings from quadrantal bearing system so its range is from 0 to 90 degrees and measured clock n anticlock wise. Whereas Azimuth refers to readings from Whole Circle Bearing system and measured clockwise only. Just remember this.. Now here in this problem or any other...it is pretty evident from the question that we are working in WCB system. So when we say bearing...we are just referring to the observations in that problem and nothing else.
Sir,
why 90 ° 8 is in the second quadrant?
why 60° 22 in the third quadrant?
and why 69 ° 20 in the first quadrant?
To clarify this doubt, first read the question carefully what is given and second understand the difference between bearings and included angles.
Bearings represent the direction of a line with respect to North so line AB is 60 degrees from the North.... whereas the angles that you have mentioned, are all included angles i.e. angle between two consecutive lines and it has nothing to do with actual direction.
So, these angles (A, B, C & D) have nothing to do with quadrants.
Now, try to make the figure by yourself and see the result
(FYI: Here North is Magnetic North)
Thankyou so much sir!😊
@@CivilModules sir can you please explain how can we draw diagram using interior angles.
Beautiful, absolutely beautiful explanation and voice👌👌👌
Thank You so much sir for this amazing vedio
Well explained
Sir please make a video on finding bearing from exterior angles
wowww,🤗🤗
Sir, regarding station C, the angle looks like it is actually 60 degrees 22'. But if we check it from North direction it it NOT lying in first quadrant. Can you please explain?
No...It seems most of the students are not understanding the question properly. First, understand the difference between a bearing and an included angle.
Bearing is the angle taken from the North direction whereas an included angle is simply an angle between two consecutive lines inside of a closed polygon.
So, here AB 60 degrees, is the angle from the North whereas 60 degrees and 22' is just an angle between two consecutive lines that has nothing to do with North direction.
So, don't associate the word "Quadrant" with included angles rather use these included angles to determine in which Quadrant a particular line lies.
@@CivilModules Got it sir. Thanks for your reply. Thanks a ton! 🙏
Sir, what if excluded angles are given in question. Should subtract the excluded angle from 360 to get the included angle?
Thank you sir for concept clear
Thanks mate, I think I got the idea just wish you done this in English also
Good work
in this example we subtract angles to find FB when we add angles to find fore bearing?
Down to earth video 👍
Sir how to distribute error in bearing if there is a error in total included angle. Can we distribute the error in first bearing or only on the other three bearings.
If sum of all the included angles is not equal to (2n-4)*90 where n is number of sides, then find the error and divide that error with number of included angles. After dividing, distribute the error in each included angle. Remember, close the traverse first and then fix the bearings.
@@CivilModules Thank you Sir👍
Thanks sir ☺️
I have studied the approach in which we add the internal angle to BB to get FB and you are subtracting it. Please Clearify!
Nice👍
Greetings Sir,
Can you give the examples with already written answers.So we can match them.It clears all the doubts.Thanks.
Ye angls kis hisab se difine kiye h
sir how to calculate it as azimuth? its going to an open skitch or closed? imean for point B to C will be draw 3rd point from 2nd point to 3rd or from first to another 2nd point down? sorry for bad english i hope u answer soon,many students watch you from kurdistan.
thanks
sir,say if i started clockwise
first comes bearing BC which is equal to bearing AB(i.e, 60°) +included angle B(i.e,90°8')=150°8' but we got bearing BC=149°52' doesn't seem right.......plzz tell me how to calculate bearings if we proceed clockwise
#Shalom Christie ...sorry for the late reply. You wrote "bearing BC which is equal to bearing AB(i.e, 60°) +included angle B(i.e,90°8')=150°8'" . I'm afraid this is an incorrect approach. I strongly suggest you go through the entire compass surveying playlist and all your doubts will be cleared.
do coordinates k reference lekar angle distance se kaise nikale co-ordinate aage wala ka??
The angle C technically lies in first quadrant but why you are joining it in 4th quadrent?
Because angle C is not a bearing, it is an included angle.
Sir we have to start in anti-clockwise
You can start from either direction ! Your bearings will be same...whatever direction you choose, make sure that at the end (as a check) the bearing of line AB must be 60 degrees.
@@CivilModules TQ sir for explanation
By subtracting 90°8'from 240°then you got 149°52' how this answer came please explain
What if we got -ve FB ??
How will you get -ve FB ? Give me an example..
A wala 140 Ka Kai Karna ha sir
Hmm deside kaise krenge ki kis angle ko kitna digree le
Sir 140 2nd quardinate me na hoga
140 degree 10' is an interior angle, add it with 60 degree (FB) which gives 210 degree 10', hence 3rd quadrant.
Tnx sir
thanks sir
140 degree 2nd codrnt mn hota hai
Tq and super sir
The bearing of line AB is 133°30' and the angle ABC is 120°32' what is the bearing of BC ?
Sir please yeh karao na...
Video clear dikhai nai de rha hai
140°0' to second quadrant m ni ayega
Then I which quadrant
A baby with respect to AB
Why are you’re bearings greater than 90 degrees. I thought bearings had to be less than 90
No..not necessary. I will suggest you to watch video 4.1
@@CivilModules aren’t whole circle bearings just Azimuths? I’m just confused because my professor said if our bearings are larger than 90 then it’s wrong.
No..no need to be confused. Yes.. Technically Bearing refers to the readings from quadrantal bearing system so its range is from 0 to 90 degrees and measured clock n anticlock wise. Whereas Azimuth refers to readings from Whole Circle Bearing system and measured clockwise only. Just remember this..
Now here in this problem or any other...it is pretty evident from the question that we are working in WCB system. So when we say bearing...we are just referring to the observations in that problem and nothing else.
Koi btaiga Kya .....? Y 90'8' kaisa 2 qurardant m AA gya please btaiya ...........😭😭😭😭😭😭
Q ki first quad. Me only 90°tak angle hota hai aur yr 8' jada hai isiliye...
Sir i am confused in 180°30'
I mean angel value 180° but what is the meaning of this 30'
It means 30 minutes..and if I write 30" ..It means 30 seconds
ye value aa kaha se rahe hai ye bataya hi nahi, bekar video itna acha lecture ke bad bhi.