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 1 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)
Kuldeep nautiyal said:
8 years ago
Bearing of BC = bearing of BA + angle B.
B of BC = 30+180+120.
Bearing of CD = bearing of CB + angle C.
Bearing of CD = (330 - 180) + 60.
Angle B + angle A = 180 (sum of consecutive angle of a parallelogram are 180).
B of BC = 30+180+120.
Bearing of CD = bearing of CB + angle C.
Bearing of CD = (330 - 180) + 60.
Angle B + angle A = 180 (sum of consecutive angle of a parallelogram are 180).
(7)
Sandip mondal said:
8 years ago
<ABC=180°-60°=120°.
FB of AB=30°
BB of AB=30°+180°=210°
FB of BC=210°+<ABC=210°+120°=330°
BB of BC=330-180=150
FB of CD=150+<BAD=150+60=210°.
FB of AB=30°
BB of AB=30°+180°=210°
FB of BC=210°+<ABC=210°+120°=330°
BB of BC=330-180=150
FB of CD=150+<BAD=150+60=210°.
(1)
CHANDAN KUMAR SINGH said:
8 years ago
ANSWER-- C
Bearing of AB =Bearing of DC BECAUSE regular parallelogram.
Bearing of DC = 30 DEGREE.
NOW bearing of CD = BB of DC =180+30=210 DEGREE.
Bearing of AB =Bearing of DC BECAUSE regular parallelogram.
Bearing of DC = 30 DEGREE.
NOW bearing of CD = BB of DC =180+30=210 DEGREE.
(11)
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)
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*).
Manatosh said:
10 years ago
FB of line CD = BB of line AB.
= FB of AB + 180°.
= 30°+180° = 210°.
= FB of AB + 180°.
= 30°+180° = 210°.
(2)
Prepared said:
7 years ago
210°-180°=30°+30°=60°.
So 210° is the right answer.
So 210° is the right answer.
(4)
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.
Tushar Methiya said:
6 years ago
Bearing of AB is 30 & BAD is 60 then for CD angle is 180+30 = 210.
(10)
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