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Edexcel AS Active book question

Can anyone help me to solve this question..its on page no. 27, Q5

There are 6283 milliradians in a complete circle. The army's rounding of this to 6400 mils causes an error.

a) How far sideways from the target could this roundin cause an artillery shell to be when aimed at a target 20km away?

b) Why does the actul bearing to the target not affect your answer to a?


PS - Its u A2 not AS..my fault! sorry!
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I'm struggling with this question too.
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Kyred
I'm struggling with this question too.



Damn no one to help us out!
When you're dealing with a problem like this always try to simplify. Let's solve the problem with smaller numbers and radians instead of milliradians, which we are more familiar with.

Pretend there were only 3 radians in a circle and the army rounded to 4. A good thing here would be to draw 2 circles, one divided into 3 equal parts and the other into 4. What do you observe?
Each 'radian' in the first circle is 1/3 of the circle, so 1 'rad' = 2pi/3 actual radians.
Similarly each 'radian' in the second circle equals 1/4 of the circle, therefore 1 'army rad' = 2pi/4 = pi/2 radians.
Now draw a right triangle and label one of the acute angles theta, the leg opposite to it x and the bottom leg 20km.

In the first case, x = 20*10^(3)*sin(2pi/3) = 17 321 (5 sf)
In the second case, x = 20*10^(3)sin(pi/2) = 20 000
So the error is 20 000 - 17 321 = 2 700 (2 sf)

Now that we've understood how this works on a small scale, let's do the actual problem.

When the circle is divided into 6283 radians,
x = 20 000 sin(2pi/6283) = 20.00058383 m

When it is divided into 6400 radians,
x = 20 000 sin(2pi/6400) = 19.63495093 m

delta x = (20.00058383 - 19.63495093) m = 36.6 cm (3sf)

You can check the markscheme, this is the right answer.
There is probably more than one way to solve this.

The actual bearing to the target will not affect our answer since the error will be proportionately the same (think about it)

Hope this helps!
Original post by Ilona Basset
When you're dealing with a problem like this always try to simplify. Let's solve the problem with smaller numbers and radians instead of milliradians, which we are more familiar with.

Pretend there were only 3 radians in a circle and the army rounded to 4. A good thing here would be to draw 2 circles, one divided into 3 equal parts and the other into 4. What do you observe?
Each 'radian' in the first circle is 1/3 of the circle, so 1 'rad' = 2pi/3 actual radians.
Similarly each 'radian' in the second circle equals 1/4 of the circle, therefore 1 'army rad' = 2pi/4 = pi/2 radians.
Now draw a right triangle and label one of the acute angles theta, the leg opposite to it x and the bottom leg 20km.

In the first case, x = 20*10^(3)*sin(2pi/3) = 17 321 (5 sf)
In the second case, x = 20*10^(3)sin(pi/2) = 20 000
So the error is 20 000 - 17 321 = 2 700 (2 sf)

Now that we've understood how this works on a small scale, let's do the actual problem.

When the circle is divided into 6283 radians,
x = 20 000 sin(2pi/6283) = 20.00058383 m

When it is divided into 6400 radians,
x = 20 000 sin(2pi/6400) = 19.63495093 m

delta x = (20.00058383 - 19.63495093) m = 36.6 cm (3sf)

You can check the markscheme, this is the right answer.
There is probably more than one way to solve this.

The actual bearing to the target will not affect our answer since the error will be proportionately the same (think about it)

Hope this helps!


Please note that the question was asked in 2009. OP may no longer need any help now. It would be better that you help to answer more recent queries.
Original post by Ilona Basset
When you're dealing with a problem like this always try to simplify. Let's solve the problem with smaller numbers and radians instead of milliradians, which we are more familiar with.

Pretend there were only 3 radians in a circle and the army rounded to 4. A good thing here would be to draw 2 circles, one divided into 3 equal parts and the other into 4. What do you observe?
Each 'radian' in the first circle is 1/3 of the circle, so 1 'rad' = 2pi/3 actual radians.
Similarly each 'radian' in the second circle equals 1/4 of the circle, therefore 1 'army rad' = 2pi/4 = pi/2 radians.
Now draw a right triangle and label one of the acute angles theta, the leg opposite to it x and the bottom leg 20km.

In the first case, x = 20*10^(3)*sin(2pi/3) = 17 321 (5 sf)
In the second case, x = 20*10^(3)sin(pi/2) = 20 000
So the error is 20 000 - 17 321 = 2 700 (2 sf)

Now that we've understood how this works on a small scale, let's do the actual problem.

When the circle is divided into 6283 radians,
x = 20 000 sin(2pi/6283) = 20.00058383 m

When it is divided into 6400 radians,
x = 20 000 sin(2pi/6400) = 19.63495093 m

delta x = (20.00058383 - 19.63495093) m = 36.6 cm (3sf)

You can check the markscheme, this is the right answer.
There is probably more than one way to solve this.

The actual bearing to the target will not affect our answer since the error will be proportionately the same (think about it)

Hope this helps!


Welcome to TSR. Appreciate your effort in helping. Thanks. :smile: