Download Aptitude Crack
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Surveying - Engineering

Q1:

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

A 90°

B 120°

C 210°

D 270°

ANS:C - 210°

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). <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°. 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.