A car travels 32 km due north and then 46 km in a direction 40° west of
north. Find the DIRECTION of the car's
resultant vector.

Ifa,bandcare the

sides opposite to anglesA,B,&C.

By Cosine Rule,

b  

2

=32  

2

+46  

2

−(2×32×46cos140°)

b  

2

=1024+2116−2944cos140°

b  

2

=3140−2944cos140°

b  

2

=3140+2255.234841

b  

2

=5395.234841

b=73.45

By Sine Rule,

sinC

32

​

=  

sin140°

73.45

​

sinC=  

73.45

32sin140°

​

sinC=0.2800

C=16.26°

The direction of the resultant

vector is measured from0°east

to the resultant, this angle isθ−90°,

90°is the total angle in

the first quadrant.

∴θ−90°=C+ϕ

ϕ=50°{Alternate angles are equal}

∴The direction of the car’s

resultant vector is50°+16.26°=66.26°south of east

Step-by-step explanation:

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this