help with hard trig question: (1 Viewer)

[raven2000]

Actually three of them. The first one is the only hard one.
1- A sector of a circle with radius 5cm and an angle of pie(how do you spell this)/3 subtended at the centre is cut out of cardboard. It is then curved around to form a cone. Find its exact surface area and volume.

2- Find the diameter of the sun to the nearest kilometre if its distance from the earth is 149000000 and it subtends an angle of 31'(not degrees, minutes) at the earth.

3- Given that the wingspan of an aeroplane is 30m, find the planes altitude to the nearest metre if the wingspan subtends an angle of 14'(minutes) when it is directly overhead.

Answers: 1: Surface area = 25PIE/6 cm squared.
Volume: 125root35pie OVER 648 cm cubed
2- 1343622km
3- 7367m

Note: i think that the answer to question1 might be wrong or maybe i just noobed it. I spent ages on it and never got the right answer.

 

[lyounamu]

Actually three of them. The first one is the only hard one.
1- A sector of a circle with radius 5cm and an angle of pie(how do you spell this)/3 subtended at the centre is cut out of cardboard. It is then curved around to form a cone. Find its exact surface area and volume.

2- Find the diameter of the sun to the nearest kilometre if its distance from the earth is 149000000 and it subtends an angle of 31'(not degrees, minutes) at the earth.

3- Given that the wingspan of an aeroplane is 30m, find the planes altitude to the nearest metre if the wingspan subtends an angle of 14'(minutes) when it is directly overhead.

Answers: 1: Surface area = 25PIE/6 cm squared.
Volume: 125root35pie OVER 648 cm cubed
2- 1343622km
3- 7367m

Note: i think that the answer to question1 might be wrong or maybe i just noobed it. I spent ages on it and never got the right answer.

1) A=1/2 . r^2 . @
= 1/2 . 5^2 . pi/3
= 25pi/6 cm^2

For this one, you just find the sector area. I also found the surface area of the cone but it seems to me that it isn't part of the entire surface area.

V= (sqaure root of (875/36))/3 x 25pi/36 = (25pi . square root of 875)/648
= (125pi . square root of (35))/648 cm^3
= 3.58524552... cm^3
= 3.59 cm^3 to the two decimal places.

The circumference of the circle of the cone = l of the sector = 5pi/3
Using that you will be able to find the diameter of the circle of the cone because d . pi = 5pi/3
Therefore d = 5/3
Using that, you will be able to find the surface area of the circle of the cone.

Then you use the Pythagoras' theorem to find the height of the cone like this:
(5/6)^2 + x^2 = 5^2
Therefore, x = square root of (875/36)
Then find the volume. Done!

2) Draw the right-angled triangle. Let the acute angle be 31 minutes (the smallest angle when you draw a right-angled triangle). Let the opposite angle be x (which is the diameter of the sun) and the hypotenuse is 149000000.

31 minutes = 31/60 degree = 31/60 . pi/180 = 0.00901753446...

sin 31 minutes = x/149000000
x = sin 0.00901753446... times 149000000
= 1343594.426 km

3) Draw the right-angled triange where the smallest triangle points upwards. Let the height be the adjacent side of the angle. You are given the opposite side.

14 minutes = 14/60 . pi/180 = 4.07243492...
tan(4.07243492...) = 30/x
x = 30/tan(4.07243492)
= 7366.559499...
= 7366.56 m to the two decimal places.

 

[lyounamu]

I just realised that my 2nd question was different from the solution. Can someone post the working out? I just want to see where I may have made a mistake.

 

[raven2000]

thanks alot for question 1 and 3
Dont worry about question 2, i justed wanted to know, but its not going to be in my exam this term.
About question 1, isn't it supposed to be the surface area. argh.. i spent ages on it, and i still kept getting the wrong answer, because they asked the wrong question?.. ..

PS: are u in year 11 (hsc 2009)

 

[Aplus]

Yeah, he is in Year 11, but accelerated in Mathematics Extension 1.

 

[lyounamu]

thanks alot for question 1 and 3
Dont worry about question 2, i justed wanted to know, but its not going to be in my exam this term.
About question 1, isn't it supposed to be the surface area. argh.. i spent ages on it, and i still kept getting the wrong answer, because they asked the wrong question?.. ..

PS: are u in year 11 (hsc 2009)

Well. In questin 1, I worked out the SA of the cone (with out the bottom of the cone itself). So I basically worked out the SA of the sector by using the formula: A = 1/2 . r^2 . @ (where @ is in radian).

Yeah, I am in Year 11 doing HSC Mathematics and Mathematics Extension 1.

 

[jellybelly59]

omg i did question 1 at tutor.... Maths in focus LOL i kept getting that wrong too... cause i didn't notice it was an open cone >.<"

 

[lacklustre]

Rather than create my own thread, I'll just post a question here:

A circle has chord of 25mm with an angle of pi/6 subtended at the centre. Find, to 1 decimal place, the length of the arc cut off by the chord.

 

[lyounamu]

Rather than create my own thread, I'll just post a question here:

A circle has chord of 25mm with an angle of pi/6 subtended at the centre. Find, to 1 decimal place, the length of the arc cut off by the chord.

Let radius = r

If the one angle is pi/6, other two angles are 5pi/12 (75 degrees) because they are equal since the triangle is an isoceles triangle with two sides equal (which are two radii).

Using sine rule:

25/(sin pi/6) = r/(sin 5pi/12)
r = 9.81197428...

Now, l = r@
= 9.811974328... x pi/6
= 5.1375377...
= 5.1mm to the first decimal place.

 

[lacklustre]

Let radius = r

If the one angle is pi/6, other two angles are 5pi/12 (75 degrees) because they are equal since the triangle is an isoceles triangle with two sides equal (which are two radii).

Using sine rule:

25/(sin pi/6) = r/(sin 5pi/12)
r = 9.81197428...

Now, l = r@
= 9.811974328... x pi/6
= 5.1375377...
= 5.1mm to the first decimal place.

Cheers!

 

[lacklustre]

Could anybody tell me how to approach these types of questions? (trig functions):

e.g. lim(x --> 0) sin3x/5x or lim(x-->0) 1/x sin(x/2pi)

Tbh i don't know how to do any of these types so could someone explain it to me?(i went through the textbook, but got confused).

 

[lyounamu]

Could anybody tell me how to approach these types of questions? (trig functions):

e.g. lim(x --> 0) sin3x/5x or lim(x-->0) 1/x sin(x/2pi)

Tbh i don't know how to do any of these types so could someone explain it to me?(i went through the textbook, but got confused).

lim (x --> 0) sin3x/5x = lim (x--> 0 )sin3x/3x . 3/5
= 1 . 3/5 because lim (x --> 0) sin 3x/3x = 1 (rule that you have to know)
= 3/5

lim (x--> 0) 1/x sin(x/2pi) = 0 because 1/x approaches 0 when x becomes too big.

 

[lacklustre]

lim (x --> 0) sin3x/5x = lim (x--> 0 )sin3x/3x . 3/5
= 1 . 3/5 because lim (x --> 0) sin 3x/3x = 1 (rule that you have to know)
= 3/5

lim (x--> 0) 1/x sin(x/2pi) = 0 because 1/x approaches 0 when x becomes too big.

could you explain what you did there? (the bold part)

 

[lyounamu]

could you explain what you did there? (the bold part)

What I did was I changed the denominator 5x to 3x by putting 3/5 in. By putting 3/5 in I made the denominator equal to the angle of the sin so I can say in the end that the sin 3x/3x = 1.

 

[lacklustre]

What I did was I changed the denominator 5x to 3x by putting 3/5 in. By putting 3/5 in I made the denominator equal to the angle of the sin so I can say in the end that the sin 3x/3x = 1.

Cheers Iyounamu, makes perfect sense now.

 

[lacklustre]

Again I am stuck (what's new?):

I need to graph y = cos(2x - pi/4)

I know it's y=cos2x shifted to the right by pi/8. I just don't know the technique to graph it.

Also i know amplitude is 1; period (T) = pi ... yet i still can't graph it.

EDIT:While I'm here, I might also ask how one would graph something like y = 2cosx + 3sinx

 

[sicmacao]

Again I am stuck (what's new?):

I need to graph y = cos(2x - pi/4)

I know it's y=cos2x shifted to the right by pi/8. I just don't know the technique to graph it.

Also i know amplitude is 1; period (T) = pi ... yet i still can't graph it.

EDIT:While I'm here, I might also ask how one would graph something like y = 2cosx + 3sinx

to graph y = 2cosx + 3sinx, use the formula 2cosx + 3sinx = rcos(x-b)

where r = sqrt(2^2+3^2) = sqrt(13)
and b = tan^-1(3/2)

graphing y = sqrt(13) cos(x-b) involve shifting cos graph b units to the right and stretch the amplitude by sqrt(13).

 

[VenomP]

Let radius = r

If the one angle is pi/6, other two angles are 5pi/12 (75 degrees) because they are equal since the triangle is an isoceles triangle with two sides equal (which are two radii).

Using sine rule:

25/(sin pi/6) = r/(sin 5pi/12)
r = 9.81197428...

Now, l = r@
= 9.811974328... x pi/6
= 5.1375377...
= 5.1mm to the first decimal place.

When I do the bolded part, I always get 48.29. Can you explain how you get 9.81 from it?

25/(sin pi/6) = 50, isnt it?

And (sin 5pi/12) = 0.9659... right?

I don't get how you came out with 9.8.

 

[tommykins]

25sin(75)/sin(30) = 48.3, not 9.8

 

[lyounamu]

I just realised that. I probably didn't change the mode of the calculator at that time.