more math...

[perfectspring]

Hopefully the last math questions i have before school actually starts :-P

Comments

[moesbarn]

for #3 i got 796 ft. i dont really feel like explaining it tho, too tired... ill come back and attempt the others later

[darth_phoenix]

I disagree! I drew a line from the top of the tower perpendicular to the height of the plane, making the triangle with 90º and 2000 ft leg. The angle adjacent to the 2000 ft leg must be 62º, since it sums to 90º with the 28º angle of depression. Thus the distance from the plane at A to the top of the tower is 2000 Cos[62º]. We know that the "left" angle at B is 34º, so the "right" angle is going to be 146º since the two sides sum to 180º. The final angle of the AB(tower) triangle is 6º, since 6º+146º+28º=180º. Thus using the sine rule, and labeling the distance |AB| as x, we get:

x/Sin[6º] = 2000 Cos[62º]/Sin[146º]
x=176 ft to the nearest foot (it's actually around 175.514)

[moesbarn]

i stand by my original answer, i didnt think of using the sine rule, i did it the long way. but even if u do use the sin rule, it would be 2000/cos62 not 2000cos62. if u do it ur way the distance from A to the tower is 939 ft, which is impossible cause its 2000 ft if u just go straight down.

[darth_phoenix]

Marf. *ahem* Yes. I get 796 now. Wow I've been chumped. Twice. :) Ah well.

[moesbarn]

ok im back, i got 72 degrees on #2 as well, and im a lil bit sure about my answer than u sound :p

[doodlelicious]

1. The oblique asymptote is x+1 seeing as you simply divide the top by the bottom.
2. Geometry+ algebra= too lazy to do at `10:28 in the morning.
3. Similar question on our exam last year! use the sin/cosine rule, whichever seems convienient (use sin actually seeing as you only need one length and you know the angle must be acute)

Hopefully that helped you out!

[darth_phoenix]

If you divide the top by the bottom you get 6x-4...

(x^3+5x-4)/(x^2-x) = x+1 + (6x-4)/(x^2-x).

Check it.

[doodlelicious]

I know that, but the asymptote is the one that's the.....hmm......the number that's in it completely! The thing on top of the dividing symbol.
That's terribly comprehendable, don't you think.
6x-4 is the leftover, that's why we put it over x^2-x
and I graphed it.....and graphs win!

[darth_phoenix]

Yes, and so the oblique asymptote is 6x-4, and NOT x-1...

[darth_phoenix]

First of all, anything mathyful you look up FIRST on mathworld, and then if it's not there (rare) or you don't understand, look elsewhere.

http://mathworld.wolfram.com/x-Axis.html

For the second question, pretend the radius of the circle is 1 (it doesn't matter, but this makes explanation easier). If arc-length s=rθ=θ, then the ratio of the arc-lengths means the ratio of the angles. If 5θ=180º, then θ=36º (the interior angle AOC). Then the angle ABC is half of the angle AOC (from silly geometry theorem), and then CBD is 90-ABC which is 90-36/2 = 90-18=72º. Which is what you got. Hooray.