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A boat is sailing on a course of 340T for a distance of 100km. It then makes a left turn of 10 degrees and travels for 50km
find the new bearing and find the distance of the boat from the home.
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A scout troop travels on a hike of 2km with a bearing of N48 E. they then turn south and travel on a bearing of S55 E and travel for 5km and camp for lunch.
How far is the camp from the starting point and find the bearing to get the scout troop back to their starting point
need help with both they have confused me greatly
Thanks!
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"A boat is sailing on a course of 340T for a distance of 100km. It then makes a left turn of 10 degrees and travels for 50km."
Unless I misunderstand your question, then new bearing will be (340-10=) 330.
I find these are easiest to do by converting to (x,y) distances first.
340 (20 W of North) for 100 km is:
100 * sin(20) = 100 * 0.342 = 34.2 km West
100 * cos(20) = 100 * 0.940 = 94.0 km North
330 (30 W of North) for 50 km is:
50 * sin(30) = 50 * 0.500 = 25.0 km West
50 * cos(30) = 50 * 0.866 = 43.3 km North
So the total of both movements are:
West: 34.2+25.0 = 59.2 km
North: 94.0+43.3 = 137.3 km
In other words, he has moved to a point that is 59.2 km West and 137.3 North of where he started
Total Distance from home (using Pythagoras) is sqrt(59.2^2 + 137.3^2) = 149.5 km
(Note: his final bearing from home (using inverse tan): atan(59.2/137.3) = 23.3 Degrees West of North = 336.7)
To be able to solve these problems, just convert from bearing/distance to x,y (ie North,East etc), do your sums, then convert back to bearing/distance.
I will leave the other one unsolved for someone else to try
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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Do you need to use trigonometry (or something) for the other one?
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
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Yes, indeed ! As the hillbilly said to the teacher "Mah boy needs as mahch trigger-nometry as he can git - he cain't shoot straight 'tall!"
I have a page which describes converting from bearing/distance (known as Polar Coordinates) to East and North (known as Cartesian Coordinates) here
I don't have a page which describes adding two bearings/distances yet. But the steps are: 1) convert to cartesian, 2) add the x's and y's, 3) convert back to polar
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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O.O I haven't learned it yet.
School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?
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Hi!
I think you have an error in the way you copied it down. I'm pretty sure it one of your degrees is slightly off. Check the problem. The way you have it, there is no right triangle.
0_o'' the way you wrote it is wrong you missed some things...