Civil Engineering - Surveying - Discussion

Discussion Forum : Surveying - Section 4 (Q.No. 1)
1.
The reduced bearing of a line is N 87° W. Its whole circle bearing is
87°
273°
93°
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
17 comments Page 1 of 2.

Vyshnavi said:   9 years ago
It is in 4 quadrant, to convert it from RB to WCB = 360-Bearing of line.

360-87 = 273.

Sir A.P.J Abdul kalam said:   9 years ago
How can we know that the bearing is in which quadrant?

Teja said:   9 years ago
I think north west is in 3rd quadrant. How you say that is in 4th quadrant?

Hari said:   9 years ago
The whole circle bearing (W.C.B) of a line is the horizontal angle measured clockwise from the North limb of the meridian.

Makvana Disha said:   8 years ago
Here, clearly given its N87°W, and North to West is 4th quadrant so the ans. is 360-87= 273.

Durge said:   7 years ago
Because in WCB method all bearing are measured north direction and in this case, it is in anticlock wise so 360 - 87 = 273.

Ben said:   7 years ago
How to know that it's anticlock wise?

Kajiji said:   7 years ago
It is in 4th quadrant, because it is northwest. North west is at 4th quadrant and it is the clockwise direction it will be 360-87=273.
(1)

Amit naik said:   7 years ago
360°-87°= 273°.

Abhishek said:   6 years ago
What if it was south east?


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