Question

The route followed by a hiker consists of three displacement vectors , , and . Vector is along a measured trail and is 2950 m in a direction 42.0 ° north of east. Vector is not along a measured trail, but the hiker uses a compass and knows that the direction is 37.0 ° east of south. Similarly, the direction of vector is 39.0 ° north of west. The hiker ends up back where she started, so the resultant displacement is zero, or + + = 0. Find the magnitudes of (a) vector and (b) vector .

Answer 1

The route followed by a hiker consists of three displacement vectors, and Vector is along a measured trail and is $2950\,\mathrm{m}$ in a direction $42.0{}^{\circ}$ north of east. Vector is not along a measured trail, but the hiker uses a compass and knows that the direction is $37.0{}^{\circ}$ east of south. Similarly, the direction of vector is $39.0{}^{\circ}$ north of west. The hiker ends up back where she started, so the resultant displacement is zero, or $= 0$. Find the magnitudes of (a) vector and (b) vector.

Solution

Let B and C be the required vectors.

Resolving each of the vectors into components N and E, the total components in each direction are:

$$E: 2950 \cos 42 + B \sin 37 - C \cos 39$$ $$N: 2950 \sin 42 - B \cos 37 + C \sin 39$$

As the total vector displacement is 0, each of these components is 0.

$$B \sin 37 - C \cos 39 = -2950 \cos 42 \quad (1)$$ $$-B \cos 37 + C \sin 39 = -2950 \sin 42 \quad (2)$$

Adding (1) $* \cos 37$ to (2) $* \sin 37$:

$$-C(\cos 39 \cos 37 - \sin 39 \sin 37) = -2950(\cos 42 \cos 37 + \sin 42 \sin 37)$$ $$C \cos(39 + 37) = 2950 \cos(37 - 42)$$ $$C = \frac{2950 * \cos(5)}{\cos(76)} = 12147,62\,\mathrm{m}.$$ $$= 3744.55\,\mathrm{m}.$$

Adding (1) $* \sin 39$ to (2) $* \cos 39$:

$$-B(\cos 39 \cos 37 - \sin 39 \sin 37) = -2950(\sin 39 \cos 42 + \cos 39 \sin 42)$$ $$-B \cos(39 + 37) = -2950 \sin(39 + 42)$$ $$B = \frac{2950 * \sin 81}{\cos(76)} = 12043,89\,\mathrm{m}.$$

Answer: (a) $12147,62\,\mathrm{m}$; (b) $12043,89\,\mathrm{m}$.