The bearings of two lines AB and AC are N90ºW and S90ºW respectively. The included angle BAC will be:
nks js[kkvksa AB vkSj AC dh fn'kk,¡ Øe'k% N90°W vkSj S90°W gSaA lfEefyr dks.k BAC gksxk%
180º
90º
0º
None of the above
0º
AB & AC are parallel to each other and in same direction.
So, angle between them is 0°.
Alternative method: Change bearing of line AB, from QB to WCB
AB = 360° – 90° = 270°
And bearing of line AC = 180° + 90° = 270°
The included equal BAC will be 270° – 270° = 0°
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