Ahiker makes four straight-line walks a 15 km at 59 ◦ b 23 km at 340 ◦ c 22 km at 125 ◦ d 30 km at 163 ◦ in random directions and lengths starting at position (41 km, 41 km), a b c d how far from the starting point is the hiker after these four legs of the hike? all angles are measured in a counter-clockwise direction from the positive x-axis.

The distance from the starting point is (-11.97 Km, 31.78 Km)

Explanation:

The easiest method for this problem is the analytical calculation of the distance traveled by the hiker, we will decompose each vector and find the result in each direction

    Va= 15 Km 59º  

    Vax= 15 cos 59= 7.73 Km

    Vay= 15 sin 59=  12.86 Km

    Vbx= 23 cos 340= 21.61 Km

    Vby= 23 sin 340= -7.87 Km

    Vcx= 22 cos 125= -12.62 Km

    Vcy= 22 sin 125 = 18.02 Km

    Vdx= 30 cos 163= -28.69 Km

    Vdy = 30 sin163 = 8.77 Km

We find each component of the total movement

    Vtx= Vax+ Vbx+ Vcx+ Vdx

    Vty = Vay+ Vby+ Vcy+Vdy

    Vtx= 7.73+ 21.61-12.62-28.69

    Vtx = -11.97 Km

    Vty= Vay+ Vby+ Vcy+ Vdy

    Vty= 12.86-7.87+ 18.02+ 8.77

    Vty= 31.78 Km

The distance from the starting point is (-11.97 Km, 31.78 Km)

To know what distance to advanced we add the total distances to the value of the initial distance

   (41Km,41Km) =   (Vox,Voy)

   Dx= Vtx +Vox

   Dy = Vty + Voy

   Dx= -11.97 + 41= 29.03 Km

   Dy = 31.78 + 41= 72.78 Km

This is the total distance advanced by the hiker

the answer is definitely c.

answer: pseudoscience