1. ## Find the distance

A ship travels 55 km on a bearing of 27 degree, then travels on a bearing of 117 degree for 140 km. Find the distance traveled from the starting point to the ending point.
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What I did is following:
sin27=55/hypotenuse
cos63=55/hypoenuse
*****>>>>these two give me 121 km
cos27=140/hypotenuse
Sin63=140/hypotenuse
*****>>>>>these two give me 157 km
But if I used the pythagorean thm, it gives me 150 km

I have seen similar problems they used sometimes, cos or sin or solve by pythagoran thm........It's confusing when to use what?

2. from your explanation I found you could just use the pythagoras theorem with 55 and 140 as the shorter sides as the turn of the ship was exactly 90 degrees, therefore

distance travelled $\displaystyle = \sqrt{55^2+140^2} = 150.42$

The 157 could've resulted in rounding error introduced from using the trig functions.

3. Originally Posted by Judi
A ship travels 55 km on a bearing of 27 degree, then travels on a bearing of 117 degree for 140 km. Find the distance traveled from the starting point to the ending point.
******************************************
What I did is following:
sin27=55/hypotenuse
cos63=55/hypoenuse
*****>>>>these two give me 121 km
cos27=140/hypotenuse
Sin63=140/hypotenuse
*****>>>>>these two give me 157 km
...

The bearings are outside of the triangle in question. Therefore your method will not provide you with a valid result.

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# a ship travels on a bearing of 27 deg and then travels on a bearing of 147 deg for 180km. find the distance of the end end trip from the straight point

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