/sci/ - Science & Math » Thread #10072275

50KiB, 802x756, cos.jpg

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Anonymous Sun 14 Oct 23:23:03 2018 No.10072275 View Reply Original Report

Quoted By:

Can someone please explain the steps and methods of doing this question, since im kinda confused!

Anonymous

Anonymous Sun 14 Oct 2018 23:32:52 No.10072289 Report

Quoted By: >>10072291 >>10072298 >>10072302 >>10072351

Let's place the hexagon inside the cartesian plane, so B=(0,0). Since BEC=15cm, the angle between B and the horizontal line is 45 cm. Then, C=(xcos(45),-xsen45). Similarly with Q=(10cosBXQ,-10senBXQ), where BXQ is the angle formed with the horizontal axis. But since Q and C have the same horizontal coordinate, 10cosBXQ=xcos(45).
Finally, BXQ=90-PBQ.

Anonymous

Anonymous Sun 14 Oct 2018 23:33:54 No.10072291 Report

Quoted By: >>10072298

>>10072289
Also, use >>>/wsr/ next time.
Or >>>/sci/sqt
wsr is honestly better desu.

Anonymous

Anonymous Sun 14 Oct 2018 23:38:20 No.10072298 Report

Quoted By:

>>10072289
>>10072291
Thank you for explaining this, i thought people on sci would be better suited to this!

Anonymous

Anonymous Sun 14 Oct 2018 23:40:32 No.10072302 Report

Quoted By:

>>10072289
Ah, small mistake, but multiply PBQ by two.

Anonymous

Anonymous Mon 15 Oct 2018 00:17:17 No.10072351 Report

Quoted By:

>>10072289
does this actually work? looks too simple to me

i got the answer first writing $EBQ = \theta$
and you can look at the picture and say
$\cos(\theta) = \frac{x\sin(15^\circ)}{10}$

then use the double angle formula to find $\cos(2\theta)=2\frac{x^2 \sin^2(15)}{10^2}-1$

half angle formula gives you
$\sin^2(15^\circ) = \frac{1-\frac{\sqrt{3}}{2}}{2}$

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