Answer to Question #147259 in Trigonometry for jared

Question

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.

Answer 1

Explanations & Calculations

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

$\qquad\qquad \begin{aligned} \small l &= \small ut =16.5kmh^{-1}\times 1h=16.5km\\ \\ \small \theta &= \small 34^040^{'}=34^0+\frac{40}{60}^0=34.667^0\\ \small \alpha &= \small 60^010^{'}=65^0+\frac{10}{60}^0=65.167^0 \end{aligned}$

• Therefore, by trigonometry

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