Question

Trigo problem #3

Answer 1

good afternoon :)

Answer 2

o.o How do you know it's afternoon?.......

Answer 3

it's morning here ... and haven't slept.

Answer 4

@experimentX You rarely sleep :P

Answer 5

how do you know?? i sleep in the afternoon.

Answer 6

o.o I remember you said something like 'it's 5:30am here and I'm planning to sleep' in a post. But about 4 hours later, you went online.

Answer 7

haha .. 10am right now.

Answer 8

So..2 hours difference. Any ideas on how to solve the problem?

Answer 9

not really ... thinking.

Answer 10

its A 15 degrees

Answer 11

It's not A

Answer 12

Sorry, breakfast first. Can't think when I'm hungry :S

Answer 13

ohh let me check again my solution

Answer 14

Back, do you need the answer first?

Answer 15

no ... i think i would like to do see this problem for now.
http://openstudy.com/study#/updates/4f87af5ee4b0505bf0871b19

Answer 16

Take your time

Answer 17

I've managed to show that it's not 15, but no more progress so far.

Answer 18

And less than 75, but that doesn't rally help us.

Answer 19

At least we can sort out one of the answers now :)

Answer 20

I'm gonna say 50 degrees.

Answer 21

Hmm, if the solution is correct, then your answer is not correct...Sorry :(

Answer 22

I have reduced the upper bound to 45. So the angle must be less than 45 degrees.

Answer 23

So it must be B or C.

Answer 24

@KingGeorge Right~ How did you do that???

Answer 25

angle DAC + angle ADE = 75 degrees ..

Answer 26

Got it

Answer 27

I extended line AB along with line CD until they met at point F. Then I extended line BC some undefined amount. Then I drew a line from F perpendicular to AB so that it intersected line BC at point G.

With some triangles, you get that angle BGF is 30 degrees. Then draw a line from G to A. The angle of BGA must be less than 30 degrees. By extending line AD, and doing some algebraic manipulation of variables, I got that our desired angle must be less than 45 degrees.

Answer 28

Namely, you can draw a line from D to G. Let angle ADE=x, angle CGF=x', and angle CDG=y.

Then we have the relations \(x+y=120\) and \(x'+y=105\). Thus, \(x-x'=15\) since \(x' < 30\), this implies \(x<45\).

Answer 29

I'm not sure how to prove it's either B or C, but if I had to guess, I would say B.

Answer 30

it's B :)

Answer 31

Given that, I would have to assume the easiest proof somehow involves proving triangle BEC is similar triangle ACD.

Answer 32

Hmm thanks... I think this time I really have to ask my teacher. Thank you so much!!!!!

Answer 33

Duh. I see an easy way now. Hold on for two minutes.

Answer 34

We know from hypothesis that BC=AC and that EC=DC. Also, it's given that angle BCE=45. Since angle ECD is 60, we know that angle ACD=45. Thus, we have a Side-Angle-Side to prove triangles BEC and ACD are congruent. Hence, angle ADC=angle BEC=95 degrees.

95=60=35

QED

Answer 35

@KingGeorge You're so smart!!!!!!!!!!!! Thank you so much!!!!!

Answer 36

You're welcome. Time for bed (for me) :)

Answer 37

Night :)

Answer 38

@KingGeorge , good job man...

Answer 39

@dpaInc agree :)