Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 1 (Q.No. 48)
48.
ABCD is a regular parallelogram plot of land whose angle BAD is 60°. If the bearing of the line AB is 30°, the bearing of CD, is
Discussion:
16 comments Page 2 of 2.
DARSHAN H A said:
10 years ago
Total interior angle = (2n-4)*90 where n=4, no of sides.
There fore total interior angle = ((2*4)-4)*90 = 360. Each included angle = total interior angle/n = 360/4 =90.
The defection angle = 180-each included angle = 180-90 = 90.
Then fore bearing of any line = Deflection angle + Fore bearing of previous line.
Fore bearing of AB = 30, fore bearing of BC = 90+30 = 120, fore bearing of CD = 90+120 = 210, fore bearing of DA = 90+210 = 300.
There fore total interior angle = ((2*4)-4)*90 = 360. Each included angle = total interior angle/n = 360/4 =90.
The defection angle = 180-each included angle = 180-90 = 90.
Then fore bearing of any line = Deflection angle + Fore bearing of previous line.
Fore bearing of AB = 30, fore bearing of BC = 90+30 = 120, fore bearing of CD = 90+120 = 210, fore bearing of DA = 90+210 = 300.
(1)
Mohan Cenation said:
1 decade ago
The bearing of AB is 30*.
Since, it is less than 90*. So, we have to add 180* for it.
Then it is (30*+180*=210*).
Since, it is less than 90*. So, we have to add 180* for it.
Then it is (30*+180*=210*).
Raj said:
1 decade ago
Back Bearing-Fore Bearing = 180.
So BB-30 = 180,
Now, BB = 180+30 = 210.
So BB-30 = 180,
Now, BB = 180+30 = 210.
Mohammed libinshad. k said:
1 decade ago
Since it is a regular pentagon.
i.e. Bearing of AB = Bearing of DC.
=> Bearing of CD = Back bearing of DC.
= Back bearing of AB = 180+30 = 210.
i.e. Bearing of AB = Bearing of DC.
=> Bearing of CD = Back bearing of DC.
= Back bearing of AB = 180+30 = 210.
(1)
Debashis Mahato said:
1 decade ago
How its possible?
HITESH KUMAR said:
1 decade ago
Bearing of CD = 180+30 = 210.
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