Answer:

150° with the 1ST vector

Explanation:

the resultant of three VECTORS 1,2 and 3 units whose directions are those of the sides of an equilateral triangle at an angle of is(a) 30 with the 1st vector (b) 15 with the 1st vector (c) 100 with the 1st vector (d) 150 with the 1st vector

Taking AB as base of the triangle and draw the two sides each making an angle of 60 degree with AB at A and B respectively meeting at the point C. Assume vectors or forces of magnitude 1 ,2 and 3 act respectively along the side AB,  BC and CA.

Here THIRD vector is CA  not AC

AB = 1 i  ( along with x axis)

BC = -2Cos60°i + 2Sin60°j  = - 1 i  + √3 j

we extend CA further from point A

CA = - 3Cos60° - 3Sin60°j  = -3/2 i - 3√3 /2 j

Net resultant AB + BC + CA  = 1 i - 1 i  -3/2 i  + √3 j - 3√3 /2 j

= -3/2 i - √3/2 j

Magnitude =  √ (-3/2)² + (-√3/2)²  =  √ (9/4 + 3/4) = √3

Both are - ve so third quadrant

Cosθ = -3/2/ √3 = -√3/2  =>  θ = 210°  ( in 3rd Quadrant)

Angle between AB & Resultant = 360 - 210  = 150°

150° with the 1st vector