Civil Engineering - Surveying - Discussion

Discussion Forum : Surveying - Section 1 (Q.No. 40)
40.
The bearings of the lines AB and BC are 146° 30' and 68° 30'. The included angle ABC is
102°
78°
45°
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
40 comments Page 4 of 4.

Satyendra patel said:   10 years ago
180-146°30' + 68°30' = 102'.

PRAVESH VERMA said:   10 years ago
Both f.b.

So BC of B.B =68°30'+180° = 248°30'.

AB of f.b. = 146°30'.

/ABC = 248°30' - 146°30' = 102° answer.

Darshan H A said:   10 years ago
Included angle = Bearing of previous line - Bearing of next line

= bearing of ba - bearing of bc

= (180+146.5)-(68.5) = 258 which is exterior angle.

Hence interior angle = 360-258 = 102.

Darshik said:   10 years ago
146.5-90 = 56.5.
90-56.5 = 33.5.
33.5+68.5 = 102.

Chandan said:   1 decade ago
180°-146°30' = 33°30'.

68°30'+33°30' = 102°.

Dhaval naam to suna hi hoga said:   1 decade ago
Simple 180-146°30'+78°30'=102.

Danesh said:   1 decade ago
Included angle = f.b of next line - b.b of previous line.
= f.b of BC - b.b of AB.
= 68.30 - 146.30.
= 78.
+180 = 102.

PRITHVEE RAJ said:   1 decade ago
INCLUDED ANGLE = 180'-146'30'+68'30' = 102'00'.

Pawan kumar said:   1 decade ago
146'30'-90' = 56'30.

90'-56'30' = 33'30+68'30 = 102'00'.

Kala said:   1 decade ago
Answer: 146-68 = 78.

78+180 = 258.

360-258 = 102.


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