Two hikers set out from camp. the first hikes 4 miles due west then turns 40° south and hikes 1.8 miles. second hikes 4 miles to E

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Two hikers set out from camp. the first hikes 4 miles due west then turns 40° south and hikes 1.8 miles. second hikes 4 miles to East then turns 52° north toward North in Heights 1.8 miles which hiker is farther from the camp explain

2 thoughts on “Two hikers set out from camp. the first hikes 4 miles due west then turns 40° south and hikes 1.8 miles. second hikes 4 miles to E”

  1. Answer:
    The second hiker is farther away from the visitor center as all things being equal cos 128 < cos 140
    from distance^2 = 4^2 + 1.8^2 — 2*4*1.8 cos (angle turned).
    Step-by-step explanation:
    brainliest pls

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  2. Answer:
    The first hiker is the farthest from camp since 140° > 128°.
    Step-by-step explanation:
    The journeys of both hikers can be modelled as triangles (see attachment).
    Even though they hike the same distance of 4 miles plus 1.8 miles, as the angles of their turns between the two legs of their journeys are different, the final distance they are from camp is also different.
    The first hiker turns 40° south so the included angle of the triangle is 140°.
    The second hiker turns 52° north, so the included angle of the triangle is 128°.
    As 140° > 128°, the distance between the first hiker and camp is farther than the distance between the second hiker and camp.
    To prove this, use the cosine rule to find the missing length of each triangle.
    [tex]\boxed{\begin{minipage}{6 cm}\underline{Cosine Rule} \\\\$c^2=a^2+b^2-2ab \cos C$\\\\where:\\ \phantom{ww}$\bullet$ $a, b$ and $c$ are the sides.\\ \phantom{ww}$\bullet$ $C$ is the angle opposite side $c$. \\\end{minipage}}[/tex]
    First hiker:
    [tex]\implies c^2=4^2+1.8^2-2(4)(1.8) \cos 140^{\circ}[/tex]
    [tex]\implies c^2=19.24-14.4\cos 140^{\circ}[/tex]
    [tex]\implies c=\sqrt{19.24-14.4\cos 140^{\circ}}[/tex]
    [tex]\implies c=5.501912393[/tex]
    Second hiker:
    [tex]\implies c^2=4^2+1.8^2-2(4)(1.8) \cos 128^{\circ}[/tex]
    [tex]\implies c^2=19.24-14.4\cos 128^{\circ}[/tex]
    [tex]\implies c=\sqrt{19.24-14.4\cos 128^{\circ}}[/tex]
    [tex]\implies c=5.301464443[/tex]
    As 5.50 > 5.30, this proves that the first hiker is farthest from the camp.

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