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/sci/ - Science & Math

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2648251 No.2648251 [Reply] [Original]

The picture represents a circular courtyard surrounded by a high stone wall. A flood light, located at e, shines into the courtyard. A person walks from the center c alon cb to b at a rate of 6 ft/sec. The radius of the courtyard is 60ft.

How far has the person shadow traveled in 2 seconds?

also what is the general formula?

>> No.2648273

bump for help

>> No.2648295
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2648295

>> No.2648306
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2648306

>>2648251

23.7 metres

now kindly get the fuck out of my /sci/!

>> No.2648315

>>2648306
metres? the question is in feet your fucking dumbshit

>> No.2648319
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2648319

just some hints op, maybe I come back here later:

r:=ec
x = v·t
tan(bla)=x/r

also D is linear in the other angle

make another pic where you name the other quantitier, then it's easier to talk

>> No.2648336

here you go op

2arctan(t/10)*60=D

>> No.2648347

>>2648306
why you mad?

>> No.2648361

>>2648315
lol, oh yeah.

feet then.

and fuck you ya white honkey nigger!

>>2648336
well played asshat, did OP ask for the fucking diamtere???

NOOOO!!!

so shut the fuck up!

>> No.2648364

>>2648347
Because everyone hates him.

>> No.2648370

>>2648364
*her

>> No.2648378

>>2648361
why you such a bitch ek?

>> No.2648451

>>2648378
because you touch you're-self at night.

>> No.2648465

>>2648370
*him

>> No.2648468

>>2648465
What?
Ek is a girl right?

>> No.2648508
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2648508

Find angle ECD
This may require the law of sines.

>σνμβολα aphiski

>> No.2648623
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2648623

>>2648508
>ECD

fullretard

E and C are points, while D is a distance on the circumference of the circle

you fucking moron!

>> No.2648719
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2648719

Using some geometry, you can figure out that

<div class="math">\theta =arctan(\frac{x}{60})-arctan(\frac{60}{x})+\frac{\pi }{2}</div>

The length of arc D is <span class="math">D=60 \theta[/spoiler]

The position x is <span class="math">x=6t[/spoiler]. After two seconds, <span class="math">x=6(2)=12[/spoiler]

Plug this into the equation for D:

<div class="math">D =60 \left [ arctan(\frac{12}{60})-arctan(\frac{60}{12})+\frac{\pi }{2}\right ]\approx 23.687</div>

So the shadow moves about 23.687 ft.

>> No.2648741

>>2648719
tl;dr: answer is exactly what EK said it was at the start of the thread.
EK is never wrong. >>2648306