# Bmat maths helpWatch

#1
Hi all!!

I have been trying to solve this question for a while now, but unfortunately ....

I know that you always have to look which point is the 'from' point. If you know the 'from' point, you are supposed to draw a straight line pointing to the north and then the angle clockwise is the angle/bearing. However, this question is so weird, I do not know which is the 'from' point and how to draw and solve this question.

19 I stop while out walking and take the bearing of a windmill and note it as θ º. I then walk 5 km
north and take the bearing again – it is now 2 θ.
How far away, in km, was the windmill from the position where I took the first bearing?
A 5 cos θ
B 5 cos 2 θ
C 10 cos θ
D 5 sin 2 θ
E 10 sin θ

Thanks a lot!
0
4 years ago
#2
(Original post by Roqia)
Hi all!!

I have been trying to solve this question for a while now, but unfortunately ....

I know that you always have to look which point is the 'from' point. If you know the 'from' point, you are supposed to draw a straight line pointing to the north and then the angle clockwise is the angle/bearing. However, this question is so weird, I do not know which is the 'from' point and how to draw and solve this question.

19 I stop while out walking and take the bearing of a windmill and note it as θ º. I then walk 5 km
north and take the bearing again – it is now 2 θ.
How far away, in km, was the windmill from the position where I took the first bearing?
A 5 cos θ
B 5 cos 2 θ
C 10 cos θ
D 5 sin 2 θ
E 10 sin θ

Thanks a lot!

Draw a diagram taking an acute angle of about 30 degrees as theta. Draw W somehere along this line. Then go due north until the angle to W is doubled. You need the exterior angle of a triangle rule to show what sort of triangle you have. The answer then drops out with simple trig.
0
#3
(Original post by unclefred)
Draw a diagram taking an acute angle of about 30 degrees as theta. Draw W somehere along this line. Then go due north until the angle to W is doubled. You need the exterior angle of a triangle rule to show what sort of triangle you have. The answer then drops out with simple trig.
I am sorry, but I do not understand where I have to draw W (along which line?).
0
4 years ago
#4
(Original post by Roqia)
I am sorry, but I do not understand where I have to draw W (along which line?).
anywhere you like along the line from your startpoint O on a bearing of theta (make it look about 30 degrees)
0
4 years ago
#5
(Original post by Roqia)
I am sorry, but I do not understand where I have to draw W (along which line?).

Try this. Mark a point O on paper. Now draw a line at approx bearing of 30 degrees. Make it a long line. Now using a scale of 1cm = 1 km draw a pony 5 cm north of O. From this point draw a line on a bearing of 2 theta (about 60 degrees. W is where the lines intersect. Now ignore scale and physical angles. Mark one side of the triangle as 5km and the two angles of theta and 2 theta and do the trig mentioned in my previous post. It is important you recognise the sort of triangle you've got.
0
#6
(Original post by unclefred)
Try this. Mark a point O on paper. Now draw a line at approx bearing of 30 degrees. Make it a long line. Now using a scale of 1cm = 1 km draw a pony 5 cm north of O. From this point draw a line on a bearing of 2 theta (about 60 degrees. W is where the lines intersect. Now ignore scale and physical angles. Mark one side of the triangle as 5km and the two angles of theta and 2 theta and do the trig mentioned in my previous post. It is important you recognise the sort of triangle you've got.
Thank you so much for your efforts, but I still have one question :

in order to draw an angle, you need two lines right? So how can I draw a line at bearing 30 degrees with only one line?
0
#7
(Original post by unclefred)
Try this. Mark a point O on paper. Now draw a line at approx bearing of 30 degrees. Make it a long line. Now using a scale of 1cm = 1 km draw a pony 5 cm north of O. From this point draw a line on a bearing of 2 theta (about 60 degrees. W is where the lines intersect. Now ignore scale and physical angles. Mark one side of the triangle as 5km and the two angles of theta and 2 theta and do the trig mentioned in my previous post. It is important you recognise the sort of triangle you've got.

I GOT IT!!!! but my only problem now is, that how am i Supposed to calculate the very long side, if I only have one side (5km) and actually only one angle (theta, because the 2 theta is an outer angle)
0
4 years ago
#8
(Original post by Roqia)
I GOT IT!!!! but my only problem now is, that how am i Supposed to calculate the very long side, if I only have one side (5km) and actually only one angle (theta, because the 2 theta is an outer angle)
Do you know the geometric rule about the exterior andgle of a triangle being equal to the sum of the two ......... This makes your triangle a well known one that all year 7's have heard of but can't spell!!
0
#9
(Original post by unclefred)
Do you know the geometric rule about the exterior andgle of a triangle being equal to the sum of the two ......... This makes your triangle a well known one that all year 7's have heard of but can't spell!!
Yes I know that rule, but how should I use it
0
4 years ago
#10
(Original post by Roqia)
Yes I know that rule, but how should I use it
OK. Lets call the start point O and the second point P. The windmill is W. The triangle has internal angle theta at O and external angle 2 theta at P. So what does this make the internal angle at W? What sort of triangle have you got? (a well known one that year 7's can't spell!!!).Can you split it into two smaller triangles each with a right angle? If you can the trig is straightforward.
0
#11
(Original post by unclefred)
OK. Lets call the start point O and the second point P. The windmill is W. The triangle has internal angle theta at O and external angle 2 theta at P. So what does this make the internal angle at W? What sort of triangle have you got? (a well known one that year 7's can't spell!!!).Can you split it into two smaller triangles each with a right angle? If you can the trig is straightforward.
I'm very sorry, but what do you mean by saying 'a well known one that year 7's can't spell' and the internal angle at W is the internal angle at W + theta, but I only know, that the internal at W is internal at B + theta, so actually that doesn't get me any further
0
4 years ago
#12
(Original post by Roqia)
I'm very sorry, but what do you mean by saying 'a well known one that year 7's can't spell' and the internal angle at W is the internal angle at W + theta, but I only know, that the internal at W is internal at B + theta, so actually that doesn't get me any further
We have a triangle OPW where OP is 5km ,the internal angle at O is theta and the external angle at P is 2 theta. So the internal angle at W must be....... Therefore the triangle OPW is an .......... triangle. (Clue is the use of 'an' rather than 'a')
0
#13
(Original post by unclefred)
We have a triangle OPW where OP is 5km ,the internal angle at O is theta and the external angle at P is 2 theta. So the internal angle at W must be....... Therefore the triangle OPW is an .......... triangle. (Clue is the use of 'an' rather than 'a')
Could you please fill in the gaps I'm Dutch and didn't lear these bearing things.. that's why I am still stuck. I also don't know yet all the ENglish words for different kind of triangles, thath's why I still don't have the answer
0
4 years ago
#14
(Original post by Roqia)
Could you please fill in the gaps I'm Dutch and didn't lear these bearing things.. that's why I am still stuck. I also don't know yet all the ENglish words for different kind of triangles, thath's why I still don't have the answer
That makes a difference. Perhaps I can bend the rules a bit and give you a bit more help than is usually allowed.

Bearings are measured clockwise from north. Begin be drawing a point O and a line up the page from O to represent North. Theta can be any angle but for simplicity of drawing choose about 30 degrees. Draw a line from O at 30 degrees to your north line. The windmill W is somewhere on this line but you don't as yet know where. Think of this practically - you can see the windmill and take a bearing but can only guess the distance. You now walk 5km north - this you CAN measure with modern gps devices. Draw a line say 5cm long and mark your second point P. From here your bearing to W is 2 theta - draw a line at 60 degrees. Where the two lines cross is you position of W.

The geometric rule you need is that the external angle of a triangle is equal to the sum of the two interior opposite ones. The external angle at P is 2theta, the internal angle at O is theta so the angle at W must also be theta. This means that the triangle OPW is isosceles (the word year 7's can't spell) and PO = PW. To make a right angle, draw a line from P to the midpoint of OW and you have two right angled triangles. Use simple trigonometry on these triangles and you have your answer.

I chose 5cm and 30 degrees purely to get a straightforward diagram, they needn't be accurate.

When you have got the answer and are satisfied that you understand it, try proving the geometrical result used above - it's a very easy proof.
2
#15
(Original post by unclefred)
That makes a difference. Perhaps I can bend the rules a bit and give you a bit more help than is usually allowed.

Bearings are measured clockwise from north. Begin be drawing a point O and a line up the page from O to represent North. Theta can be any angle but for simplicity of drawing choose about 30 degrees. Draw a line from O at 30 degrees to your north line. The windmill W is somewhere on this line but you don't as yet know where. Think of this practically - you can see the windmill and take a bearing but can only guess the distance. You now walk 5km north - this you CAN measure with modern gps devices. Draw a line say 5cm long and mark your second point P. From here your bearing to W is 2 theta - draw a line at 60 degrees. Where the two lines cross is you position of W.

The geometric rule you need is that the external angle of a triangle is equal to the sum of the two interior opposite ones. The external angle at P is 2theta, the internal angle at O is theta so the angle at W must also be theta. This means that the triangle OPW is isosceles (the word year 7's can't spell) and PO = PW. To make a right angle, draw a line from P to the midpoint of OW and you have two right angled triangles. Use simple trigonometry on these triangles and you have your answer.

I chose 5cm and 30 degrees purely to get a straightforward diagram, they needn't be accurate.

When you have got the answer and are satisfied that you understand it, try proving the geometrical result used above - it's a very easy proof.

GREAT!!! i UNDERSTAND IT Thanks a lot !!!
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