Trigonometry Modelling
I have another question I'm stuck on and have literally no clue to start. It's to do with trigonometry and bearings and I am horrible at worded questions.
"A jet ski travels 200m in a straight line on a bearing of 200°, then 600m in a straight line on a bearing of 060°. Calculate the distance the jet ski must travel, to the nearest m, and the bearing on which it needs to travel to return directly to its start point."
What I've done so far is constructed the triangle, but from there I'm absolutely clueless as to what I need to do.
Once again, thank you all.
trigonometry word-problem
edited Jan 4 at 13:05
N. F. Taussig
43.6k93355
asked Jan 4 at 12:13
Jia Xuan NgJia Xuan Ng
51
New contributor
Jia Xuan Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I have another question I'm stuck on and have literally no clue to start. It's to do with trigonometry and bearings and I am horrible at worded questions.
"A jet ski travels 200m in a straight line on a bearing of 200°, then 600m in a straight line on a bearing of 060°. Calculate the distance the jet ski must travel, to the nearest m, and the bearing on which it needs to travel to return directly to its start point."
What I've done so far is constructed the triangle, but from there I'm absolutely clueless as to what I need to do.
Once again, thank you all.
trigonometry word-problem
edited Jan 4 at 13:05
N. F. Taussig
43.6k93355
asked Jan 4 at 12:13
Jia Xuan NgJia Xuan Ng
51
New contributor
Jia Xuan Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Idea: convert the lines and distances to vectors, and add 'em up! You can put the origin at the starting point.
– Matti P.
Jan 4 at 12:31
@MattiP. I'm assuming this is an IGCSE question. If it is, the unit on vectors is after the unit on trigonometry and bearings, so the OP might not be able to use your method.
– Toby Mak
Jan 4 at 13:31
add a comment |
I have another question I'm stuck on and have literally no clue to start. It's to do with trigonometry and bearings and I am horrible at worded questions.
"A jet ski travels 200m in a straight line on a bearing of 200°, then 600m in a straight line on a bearing of 060°. Calculate the distance the jet ski must travel, to the nearest m, and the bearing on which it needs to travel to return directly to its start point."
What I've done so far is constructed the triangle, but from there I'm absolutely clueless as to what I need to do.
Once again, thank you all.
trigonometry word-problem
edited Jan 4 at 13:05
N. F. Taussig
43.6k93355
asked Jan 4 at 12:13
Jia Xuan NgJia Xuan Ng
51
New contributor
Jia Xuan Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have another question I'm stuck on and have literally no clue to start. It's to do with trigonometry and bearings and I am horrible at worded questions.
"A jet ski travels 200m in a straight line on a bearing of 200°, then 600m in a straight line on a bearing of 060°. Calculate the distance the jet ski must travel, to the nearest m, and the bearing on which it needs to travel to return directly to its start point."
What I've done so far is constructed the triangle, but from there I'm absolutely clueless as to what I need to do.
Once again, thank you all.
trigonometry word-problem
trigonometry word-problem
edited Jan 4 at 13:05
N. F. Taussig
43.6k93355
asked Jan 4 at 12:13
Jia Xuan NgJia Xuan Ng
51
New contributor
Jia Xuan Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 13:05
N. F. Taussig
43.6k93355
asked Jan 4 at 12:13
Jia Xuan NgJia Xuan Ng
51
New contributor
Jia Xuan Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 4 at 13:05
N. F. Taussig
43.6k93355
edited Jan 4 at 13:05
N. F. Taussig
43.6k93355
edited Jan 4 at 13:05
N. F. Taussig
43.6k93355
43.6k93355
asked Jan 4 at 12:13
Jia Xuan NgJia Xuan Ng
51
New contributor
Jia Xuan Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 4 at 12:13
Jia Xuan NgJia Xuan Ng
51
asked Jan 4 at 12:13
Jia Xuan NgJia Xuan Ng
51
51
New contributor
Jia Xuan Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jia Xuan Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jia Xuan Ng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Idea: convert the lines and distances to vectors, and add 'em up! You can put the origin at the starting point.
– Matti P.
Jan 4 at 12:31
@MattiP. I'm assuming this is an IGCSE question. If it is, the unit on vectors is after the unit on trigonometry and bearings, so the OP might not be able to use your method.
– Toby Mak
Jan 4 at 13:31
add a comment |
Idea: convert the lines and distances to vectors, and add 'em up! You can put the origin at the starting point.
– Matti P.
Jan 4 at 12:31
@MattiP. I'm assuming this is an IGCSE question. If it is, the unit on vectors is after the unit on trigonometry and bearings, so the OP might not be able to use your method.
– Toby Mak
Jan 4 at 13:31
Idea: convert the lines and distances to vectors, and add 'em up! You can put the origin at the starting point.
– Matti P.
Jan 4 at 12:31
Idea: convert the lines and distances to vectors, and add 'em up! You can put the origin at the starting point.
– Matti P.
Jan 4 at 12:31
@MattiP. I'm assuming this is an IGCSE question. If it is, the unit on vectors is after the unit on trigonometry and bearings, so the OP might not be able to use your method.
– Toby Mak
Jan 4 at 13:31
@MattiP. I'm assuming this is an IGCSE question. If it is, the unit on vectors is after the unit on trigonometry and bearings, so the OP might not be able to use your method.
– Toby Mak
Jan 4 at 13:31
add a comment |
1 Answer
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(DB, EA, and FC are all parallel)
If only we had $angle ABC$, we could use the cosine rule to find $AC$! Well, that's exactly what we're going to do:
$$angle EAB = 160º text{(reflex angles)}$$
$$angle DBA = 20º text{(alternate angles)}$$
$$ABC = 40º$$
Now, use the cosine rule:
$$AC^2 = 200^2 + 600^2 - 2(200)(600) cos 40º$$
$$AC = sqrt{200^2 + 600^2 - 2(200)(600) cos 40º}$$
$$= 465 text{(3 s.f)}$$
Now for part $2$, the bearing of $CA$ is just $360º - angle FCA$. However, $angle DBC$ and $angle FCB$ are corresponding angles, so $angle FCB = 120º$.
If you just work out $angle ACB$ (either sine rule or cosine rule works with $AC$ known), you can get the answer.
answered Jan 4 at 13:29
Toby MakToby Mak
3,37811128
What software are you using? GeoGebra?
– Lucas Henrique
Jan 4 at 13:33
@LucasHenrique Yep. I also haven't used Paint before.
– Toby Mak
Jan 4 at 13:34
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
(DB, EA, and FC are all parallel)
If only we had $angle ABC$, we could use the cosine rule to find $AC$! Well, that's exactly what we're going to do:
$$angle EAB = 160º text{(reflex angles)}$$
$$angle DBA = 20º text{(alternate angles)}$$
$$ABC = 40º$$
Now, use the cosine rule:
$$AC^2 = 200^2 + 600^2 - 2(200)(600) cos 40º$$
$$AC = sqrt{200^2 + 600^2 - 2(200)(600) cos 40º}$$
$$= 465 text{(3 s.f)}$$
Now for part $2$, the bearing of $CA$ is just $360º - angle FCA$. However, $angle DBC$ and $angle FCB$ are corresponding angles, so $angle FCB = 120º$.
If you just work out $angle ACB$ (either sine rule or cosine rule works with $AC$ known), you can get the answer.
answered Jan 4 at 13:29
Toby MakToby Mak
3,37811128
What software are you using? GeoGebra?
– Lucas Henrique
Jan 4 at 13:33
@LucasHenrique Yep. I also haven't used Paint before.
– Toby Mak
Jan 4 at 13:34
add a comment |
(DB, EA, and FC are all parallel)
If only we had $angle ABC$, we could use the cosine rule to find $AC$! Well, that's exactly what we're going to do:
$$angle EAB = 160º text{(reflex angles)}$$
$$angle DBA = 20º text{(alternate angles)}$$
$$ABC = 40º$$
Now, use the cosine rule:
$$AC^2 = 200^2 + 600^2 - 2(200)(600) cos 40º$$
$$AC = sqrt{200^2 + 600^2 - 2(200)(600) cos 40º}$$
$$= 465 text{(3 s.f)}$$
Now for part $2$, the bearing of $CA$ is just $360º - angle FCA$. However, $angle DBC$ and $angle FCB$ are corresponding angles, so $angle FCB = 120º$.
If you just work out $angle ACB$ (either sine rule or cosine rule works with $AC$ known), you can get the answer.
answered Jan 4 at 13:29
Toby MakToby Mak
3,37811128
What software are you using? GeoGebra?
– Lucas Henrique
Jan 4 at 13:33
@LucasHenrique Yep. I also haven't used Paint before.
– Toby Mak
Jan 4 at 13:34
add a comment |
(DB, EA, and FC are all parallel)
If only we had $angle ABC$, we could use the cosine rule to find $AC$! Well, that's exactly what we're going to do:
$$angle EAB = 160º text{(reflex angles)}$$
$$angle DBA = 20º text{(alternate angles)}$$
$$ABC = 40º$$
Now, use the cosine rule:
$$AC^2 = 200^2 + 600^2 - 2(200)(600) cos 40º$$
$$AC = sqrt{200^2 + 600^2 - 2(200)(600) cos 40º}$$
$$= 465 text{(3 s.f)}$$
Now for part $2$, the bearing of $CA$ is just $360º - angle FCA$. However, $angle DBC$ and $angle FCB$ are corresponding angles, so $angle FCB = 120º$.
If you just work out $angle ACB$ (either sine rule or cosine rule works with $AC$ known), you can get the answer.
answered Jan 4 at 13:29
Toby MakToby Mak
3,37811128
(DB, EA, and FC are all parallel)
If only we had $angle ABC$, we could use the cosine rule to find $AC$! Well, that's exactly what we're going to do:
$$angle EAB = 160º text{(reflex angles)}$$
$$angle DBA = 20º text{(alternate angles)}$$
$$ABC = 40º$$
Now, use the cosine rule:
$$AC^2 = 200^2 + 600^2 - 2(200)(600) cos 40º$$
$$AC = sqrt{200^2 + 600^2 - 2(200)(600) cos 40º}$$
$$= 465 text{(3 s.f)}$$
Now for part $2$, the bearing of $CA$ is just $360º - angle FCA$. However, $angle DBC$ and $angle FCB$ are corresponding angles, so $angle FCB = 120º$.
If you just work out $angle ACB$ (either sine rule or cosine rule works with $AC$ known), you can get the answer.
answered Jan 4 at 13:29
Toby MakToby Mak
3,37811128
answered Jan 4 at 13:29
Toby MakToby Mak
3,37811128
answered Jan 4 at 13:29
Toby MakToby Mak
3,37811128
answered Jan 4 at 13:29
Toby MakToby Mak
3,37811128
3,37811128
What software are you using? GeoGebra?
– Lucas Henrique
Jan 4 at 13:33
@LucasHenrique Yep. I also haven't used Paint before.
– Toby Mak
Jan 4 at 13:34
add a comment |
What software are you using? GeoGebra?
– Lucas Henrique
Jan 4 at 13:33
@LucasHenrique Yep. I also haven't used Paint before.
– Toby Mak
Jan 4 at 13:34
What software are you using? GeoGebra?
– Lucas Henrique
Jan 4 at 13:33
What software are you using? GeoGebra?
– Lucas Henrique
Jan 4 at 13:33
@LucasHenrique Yep. I also haven't used Paint before.
– Toby Mak
Jan 4 at 13:34
@LucasHenrique Yep. I also haven't used Paint before.
– Toby Mak
Jan 4 at 13:34
add a comment |
Jia Xuan Ng is a new contributor. Be nice, and check out our Code of Conduct.
Jia Xuan Ng is a new contributor. Be nice, and check out our Code of Conduct.
Jia Xuan Ng is a new contributor. Be nice, and check out our Code of Conduct.
Jia Xuan Ng is a new contributor. Be nice, and check out our Code of Conduct.
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Idea: convert the lines and distances to vectors, and add 'em up! You can put the origin at the starting point.
– Matti P.
Jan 4 at 12:31
@MattiP. I'm assuming this is an IGCSE question. If it is, the unit on vectors is after the unit on trigonometry and bearings, so the OP might not be able to use your method.
– Toby Mak
Jan 4 at 13:31