Civil Engineering - Surveying - Discussion

40. 

The bearings of the lines AB and BC are 146° 30' and 68° 30'. The included angle ABC is

[A]. 102°
[B]. 78°
[C]. 45°
[D]. none of these.

Answer: Option A

Explanation:

No answer description available for this question.

Kala said: (Aug 7, 2013)  
Answer: 146-68 = 78.

78+180 = 258.

360-258 = 102.

Pawan Kumar said: (Nov 30, 2013)  
146'30'-90' = 56'30.

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

PRITHVEE RAJ said: (Jan 23, 2014)  
INCLUDED ANGLE = 180'-146'30'+68'30' = 102'00'.

Danesh said: (Jul 15, 2014)  
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.

Dhaval Naam To Suna Hi Hoga said: (Jul 9, 2015)  
Simple 180-146°30'+78°30'=102.

Chandan said: (Oct 4, 2015)  
180°-146°30' = 33°30'.

68°30'+33°30' = 102°.

Darshik said: (Dec 13, 2015)  
146.5-90 = 56.5.
90-56.5 = 33.5.
33.5+68.5 = 102.

Darshan H A said: (Dec 29, 2015)  
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.

PRAVESH VERMA said: (Jan 25, 2016)  
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.

Satyendra Patel said: (Feb 26, 2016)  
180-146°30' + 68°30' = 102'.

Apurba said: (May 25, 2017)  
360 - (180 + 146°30) + 68°30 = 102.

Indrajit Maiti said: (Jul 24, 2017)  
The bb of AB - fb of BC = {(180°+146°30')- 68°30'}= 258°,
Angle = (360°-258°) =102°.

Sandip Mondal said: (Jul 31, 2017)  
FB of AB=146°30'.
BB of AB=180°+146°30'=326°30.
Ext<ABC=326°30'-BB of BC=326°30'-68*30'=258°.
Int<ABC=360°-258°=102° ans.

Pritam said: (Aug 9, 2017)  
But in book, Interior angle = BB of AB - FB of BC.

Are interior angle and included angle same.

Sura said: (Aug 13, 2017)  
FB of AB 146°30'.
B. B. of AB 146°30'+180°=326°30'.
Exterior angle ABC=326°30'-68°30'=258°.
Interior angle ABC=360°-258°=102°.

Shashi said: (Sep 7, 2017)  
Yes, right @Chandan.

Er Meghanada said: (Nov 12, 2017)  
Force bearing of AB 146°30'.

Back Bearing of AB 146°30'+180°=326°30'.

Exterior angle ABC=B.B-BC.
=326°30'-68°30'=258°.

Interior angle ABC=360°-258°=102°.

T Vijay Bhaskar said: (Dec 5, 2017)  
Included angle =180-146°30'+68°30' = 102.

Leo said: (Dec 22, 2017)  
Option d is right.

Becuase when bearing of two line measured not from their point of intersection. The rule of finding angle as different.

Leo said: (Dec 22, 2017)  
Great explanation @ Meghanada. Thanks.

Apcivilian said: (Jan 5, 2018)  
Guys take as simple.

∠ABC = F.B of BC - B.B of AB
= 68°30'- (146°30'+180°)
= -258°.
= -258° + 360°.
= 102°.

Hope it will help!

Saurabh Kumar said: (Feb 25, 2018)  
As per my knowledge, Interior angle = BB of AB - FB of BC.

Are interior angle and included angle same.

Aadarsh Devkate said: (May 30, 2018)  
180°-146°30°= 33°30°00°.

So,
<ABC = 68°30°+ 33°30°00° = 102°.

Rakesh Kumar Baral said: (Sep 4, 2018)  
AB = 107*15'.
BC= 22*00'.
AB-BC= 107*15-22*00= 85*15'.

Included angle ABC= 180*-85*15'= 94*45'.

Shramik Shanti said: (Jan 11, 2019)  
Applying formula..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.

KAN said: (Apr 6, 2019)  
BB of AB + FB of BC.
BB of AB is 180-146.30 = 33.30,
FB of BC is 68.30.
So, 33.30 + 68.30 = 102.

Mohmmed said: (Apr 14, 2019)  
The Azimuth of a line AB is 114 and the angle ABC 126°30'20". Calculate the Bearing of BC?

Please answer this.

Tarun Halder said: (Sep 6, 2019)  
Include angle b/w ABC = 108 - 146°30" + 68°30" = 102°.

Sumit said: (Apr 26, 2020)  
In finding true angle (satellite station) reduction to center, then which formula is used? Please tell me.

T.Aashish said: (Sep 6, 2020)  
Thanks for explaining @Meghanada.

Mukesh Shah said: (Oct 6, 2020)  
Includes angle = fb of next line - bb of the previous line.
= 68.30 - (146.30 + 180).
= 63.38 - 326.30.
= -258.

Angel can not be negative so it will be added by 360.
= -258 + 360.
=102.

Karan Yadav said: (Nov 26, 2020)  
180 - 146 = 34.
34 + 68 = 102.

Yojana said: (Mar 25, 2021)  
∠ABC= BB-FB.
146 30 - 68 30 = 78.
180+78= 102°.

Dipankar Barman said: (May 16, 2021)  
BB of AB - FB of BC
(180° + 146°30') - 68°30'
= 326°30'- 68°30'.
= 258°(exterior angle),

So included angle = 360°- 258°,
= 102°.

Prakash Singh Thakuri said: (Jun 26, 2021)  
(68°30' + 180°-146°30'.
= 102°00'00".

Bashant Khadka said: (Feb 20, 2022)  
180-146°30'=33°3';
Include angle = 33°3`+68°3'= 102° Ans.

Ramjan said: (Sep 11, 2022)  
Angle ABC= 360*(326°30'-68°30')
=102°00'00"
since this will be anti-clockwise so;

Interior angle =360*[(BB)AB-(FB)BC].

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