Question

Precalc problem, please help?

In the shot put, a shot must land within a 40 degree sector. Consider a coordinate system where the vertex of the sector is at the origin, one side lies along the positive x-axis, and the units are meters. If a throw lands at the point with coordinates (16, 12), how far is it from being out of bounds? [answer: 1.09m] How would I go about finding this answer?

Answer 1

Picture associated with question.

Answer 2

hmm do you know what's being asked?

Answer 3

Idk... Like what's the distance between (16, 12) and the point where it becomes out of bounds...?

Answer 4

hmmm do you know what that means?

Answer 5

the point (16, 12) is IN BOUND or in the shade of the 40 degrees angle
meaning the angle itself will encapsulate that point

so how far is that point from the BORDER line, pretty much

Answer 6

Okay... so how do I find that distance?

Answer 7

the shortest path to the BORDER line from that point, will be the PERPENDICULAR path
that is, if we go straight from that point to the BORDER line
any other line towards it, will be inclined somewhat and thus longer than that

so that'd be the shortest path, since it's the STRAIGHTEST path

Answer 8

Okay, that makes sense.

Answer 9

notice that, we'd have 2 perpendicular lines

Answer 10

so, what you really need, is to find the equation of both lines

that is, the equation of a line whose points are the origin, (0, 0) and (16, WHATEVER)
and a perpendicular line to that

to find the coordinate of the 40 degrees angle with an adjacent side of 16 units

use the \(\bf tan(\theta)=\cfrac{y}{x}\qquad tan(40^o)=\cfrac{y}{16}\implies 16tan(40^o)=y\)

Answer 11

anyhow.... you'd get with with y = 13.43 or so....

so it'd end up being the equation of a line with points at (0,0) and (16, 13.4)

that'd give you an slope

a perpendicular line to that will have, as you'd know, a "negative reciprocal" slope

and we also know it passes through the point of (16, 12)

Answer 12

once you have both equations for each line
you'd end up with a system of equations of 2 variables

if you recall, a solution for a system of equations, is where their graph meet

a solution for these 2 lines, is where they meet, well, they meet at the BORDER

that'd give a (x, y) coordinates for THAT POINT at the BORDER

how long is from there to (16, 12) ? well, just use the distance formula

Answer 13

Hmmm.... I did something wrong. I got y=13.4/16x-1.4 and y=-16/13.4x+31.1, and they meet at (16, 12)?

Answer 14

I used ... a round 13... instead of the 13.4, so I ended up with \(\bf f(x)_1=\cfrac{13}{16}x\qquad f(x)_2=-\cfrac{16}{13}x+\cfrac{60}{13}\\ \quad \\
\textit{getting rid of the denominators, I get }\\ \quad \\
f(x)_1\implies 16y=13x\implies 13x-16=0\\ \quad \\
f(x)_2\implies 13y=-16x+60\implies 16x+13y=60\)

Answer 15

Okay, but doesn't the point of intersection give you a negative y coordinate? That can't be at the border if it's a negative.

Answer 16

\(\bf 13x-16y=0\\ \quad \\
16x+13y=60\\ \quad \\ \quad \\
x=\cfrac{16y}{13}\qquad 16x+13y=60\implies 16\left(\cfrac{16y}{13}\right)+13y=60\\ \quad \\
\implies \cfrac{256y}{13}+13y=60\implies 256y+169y=780\\ \quad \\\implies 425y=780\implies y=\cfrac{780}{425}\)

Answer 17

sorry for the lag.... doing some things and ... the site got laggy too for a while

Answer 18

anyhow.... using those values..... ... the y is not negative.. but is neither a big number...
should be bigger

Answer 19

Hmmm... Well thanks for trying anyway, I think I'm just going to skip this question.

Answer 20

anyhow... I ...rewrote the equations using 13.4 this time so
this are the 2 lines -> http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIoMTMuNC8xNil4IiwiY29sb3IiOiIjRDExOTE5In0seyJ0eXBlIjowLCJlcSI6IigtMTYvMTMuNCl4Kyg0MTYuOC8xMy40KSIsImNvbG9yIjoiIzFFMThEOSJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIjEuNDM1MDAwMDAwMDAwMDA4NSIsIjI2LjgyNTYyNDk5OTk5OTk4OCIsIjMuNTYzMTI1MDAwMDAwMDAzIiwiMTkuMTg4MTI1MDAwMDAwMDAzIl19XQ--

Answer 21

so the equations are correct, as you can see, the intersect, will be in the "teens" for "x" and "y"

Answer 22

\(\bf f(x)_1=y=\cfrac{13.4}{16}x\qquad f(x)_2=y=-\cfrac{16}{13.4}x+\cfrac{416.8}{13.4}\)

Answer 23

I'm guessing something might have been off on the previous...