Question #147259
From a cruise ship travelling due north at 16.5 km/h, a wrecked ship x and an observation tower y are observed in line due east. One hour later the wrecked ship and the tower have bearings S34*40'E and S65*10'E.
Find the distance between the wrecked ship and the tower.
1
Expert's answer
2020-11-29T19:18:49-0500

Explanations & Calculations





  • Refer to the sketch attached
  • x\small x is the distance to the ship wreck from where the cruise ship was an hour ago & y\small y is the distance between wrecked ship and the light house.
  • l\small l is the distance the ship travelled during an hour &

l=ut=16.5kmh1×1h=16.5kmθ=34040=340+40600=34.6670α=60010=650+10600=65.1670\qquad\qquad \begin{aligned} \small l &= \small ut =16.5kmh^{-1}\times 1h=16.5km\\ \\ \small \theta &= \small 34^040^{'}=34^0+\frac{40}{60}^0=34.667^0\\ \small \alpha &= \small 60^010^{'}=65^0+\frac{10}{60}^0=65.167^0 \end{aligned}

  • Therefore, by trigonometry

x+y=l×tanα=35.655kmx=l×tanθ=11.411kmy=23.244km\qquad\qquad \begin{aligned} \small x+y &= \small l\times \tan\alpha=35.655km\\ \small x &= \small l\times \tan\theta=\small 11.411km\\ \small \bold{\therefore y} &= \small \bold{23.244km} \end{aligned}



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