RELATIVE VELOCITY!!!!
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wtf after the previous help., i can draw diagrams but it seems like its always wrong =[
i want to know how coem for this qtn i am wrong...(i cannot find the ans)
An aircraft is flying due south at 350km/h. The wind is blowing at 70km/h from the direction x°, where x° is acute. Given that the pilot is steering the aircraft in the direction 170°, find
value of x
the speed of aircraft in still air.
If you caculate,
70/sin 10 = 350/sin B
B = 60.25485843
Then, if you use 180 - 60 - 10 = 119 and you can find x.
but from the diagram you can already see that X is defintely less than 119 !!! how come ar??!?!?! -
1. It is already said that x is acute
2. Your diagram is half correct. Do take note of how triangular sum of forces work -
wrong mehs.
we learn
VA/W + VW = VA
x is acute ma, hence in the diagram its oso acute rite? -
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wait I read question wrongly... I check through againOriginally posted by Y_Shun:wrong mehs.
we learn
VA/W + VW = VA
x is acute ma, hence in the diagram its oso acute rite? -
heheh,
do you know that sin 60.29... is the same as sin 119.... = same answer?
because when you invert sin b to get b, it can be at any sector.
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It seems like you are correct... But I need to clarfiy something... Why is it that the question says that x is acute? Where did this question come from?
And what is the given answer? -
err...Originally posted by nightzip:heheh,
do you know that sin 60.29... is the same as sin 119.... = same answer?
TS got 1 small careless mistake:
180-60-10 = 110, not 119 -
same same, the reason I think is that when he invert sine to get angle B, the default for calculator is always the smaller angle, which is not the answer.
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hey! i'm doing this question as well and im stuck.. who knows how to do...
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Stuck 2..his diagram seemed alright...
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the wind line is supposed to give an obtuse angle lah...if you draw it according to dimensions, you will see that the angle is >90.
350 line is very long. Because sine when drawn on graph is at the positive curve from 0 deg to 180deg, of which between 0-180, there is 2 identical values for sine theta. -
yep I think this is the answer... angle B is obtuse... my mistake there...Originally posted by nightzip:same same, the reason I think is that when he invert sine to get angle B, the default for calculator is always the smaller angle, which is not the answer. -
so supposingly,
70/sin 10 = 350/sin B
B = 60.25485843; 119.7451417
but in this case how do we know which value to reject and when do we need to find the larger value? -
your question told you x is acute. That's whyOriginally posted by Y_Shun:so supposingly,
70/sin 10 = 350/sin B
B = 60.25485843; 119.7451417
but in this case how do we know which value to reject and when do we need to find the larger value? -
Anyway, if you draw a perpendicular line from the intersection of the two blue lines to the black line to form a 90deg angle at the black line side, part of the angle for angle B is already 80deg.Originally posted by Y_Shun:
If you caculate,
70/sin 10 = 350/sin B
B = 60.25485843
Then, if you use 180 - 60 - 10 = 119 and you can find x.
but from the diagram you can already see that X is defintely less than 119 !!! how come ar??!?!?!
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so we need to calculate the 2 values ar...Originally posted by eagle:your question told you x is acute. That's why
so for sine rule that we are using, we must always have 180 - basic angle? -
whats what i have on my paper, i lazy to draw out on the diagram...hence thats why i come here posting it.Originally posted by nightzip:Anyway, if you draw a perpendicular line from the intersection of the two blue lines to the black line to form a 90deg angle at the black line side, part of the angle for angle B is already 80deg. -
becauseOriginally posted by Y_Shun:so we need to calculate the 2 values ar...
so for sine rule that we are using, we must always have 180 - basic angle?
sin x = sin (180 - x)
It depends on the question whether it is acute or obtuse, or as what nightzip has said. -
This one I let Eagle answer.Originally posted by Y_Shun:so we need to calculate the 2 values ar...
so for sine rule that we are using, we must always have 180 - basic angle?
My own way is that I just observe the diagram and logic can deduce already. The reason why the question deliberately want to tell you that angle is acute is so that you will not get wrong value. -
Another QTN.
A man who swims at 1.2m/s in still water wishes to cross a river which is flowing between straight parallel banks at 2m/s. He aims downstream in a direction making an angle of 60° with the bank. Find
the speed at which he travels
the angle which his resultant velocity makes with the bank
Using cosine rule
squareroot
1.2^2 + 2^2 - 2(1.2)(2)cos60°
= 1.743559577
ans is different alot...how cums ar?
EDIT:// PAISEH, READ WRONGLY
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-.- diaoz
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so this question is settled already right?
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Originally posted by Y_Shun:
One mistake u have made. From the qn, you have said that the man aims downstream. From the diagram that u have drawn, the man is aiming upstream making an angle of 60 degrees to the bank. Instead of drawing a negative gradient line Vm/w, your Vm/w should be a positive gradient line making an angle of 60 degrees to the bank.
Another QTN.
A man who swims at 1.2m/s in still water wishes to cross a river which is flowing between straight parallel banks at 2m/s. He aims downstream in a direction making an angle of 60° with the bank. Find
the speed at which he travels
the angle which his resultant velocity makes with the bank
Using cosine rule
squareroot
1.2^2 + 2^2 - 2(1.2)(2)cos60°
= 1.743559577
ans is different alot...how cums ar?
[b]EDIT:// PAISEH, READ WRONGLY [/b]
Using cosine rule
square root
1.2^2 + 2^2 - 2(1.2)(2)cos (180 - 60) degrees
= 2.8 m/s
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Not true either. Firstly, your calculator will only give you the smallest value of the angle. Sine is positive between 0 - 180 degrees and hence this occurs. Thus, your diagram is very important in dropping you a hint whether beta is a acute or obtuse angle.Originally posted by Y_Shun:so we need to calculate the 2 values ar...
so for sine rule that we are using, we must always have 180 - basic angle?
If I use your qn as an example, if beta is an acute angle, the total angle added up will not be 180 degrees which a triangle cannot be formed since x is an acute angle. Thus, by inspection, by logic, beta angle cannot be acute. It must be obtuse. Therefore, your diagram is important as you can see. It is gd that u have drawn the diagram in such a way that x is an acute angle instead of well obtuse angle.
If the qn states that x is an obtuse angle, you will have to draw your diagram differently, drawing in such a way that x is an obtuse angle.