#I need help with this geometry question. Please help me
43 messages · Page 1 of 1 (latest)
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What can we deduct about the other angles of AXM?
I would also recommend to label angle BAD $\alpha$, and find the other angles of the parallelogram in terms of $\alpha$.
Kocher
ok i'll do that and let you know
angle BCD is 180-alpha
angle ABC + ADC =180
@pearl lily what else can i find?
are u sure abt those? perhaps if you label it on the diagram it would be easier
wait r they the same?
BCD and BAD?
yes
@wanton star notice that ABX and ADM are similar triangles.
Let AB = 1unit and AD = L
Use cosine rule in AXM to find L.
Hence use the fact that DM=MC=1/2 to find the reqd ratio
Not sure why the image is rotated btw. In my phone photos it is not
ok let me try#
@untold falcon I got the ratio 1:2
What value of L did u get
I got L=sqrt(3)
Ratio= 1/5
AM/L = AX/1
AM = L*AX
Using cosine rule
Sqrt(AX² + MX² -2cos(120)*AX*MX) = L*AX
Sub MX = AX, you get
L = sqrt(3)
Ohh
Actually I am usually not allowed to do it.
But since you specifically asked I gave it to u
WDYM?
did you figure out the error in your working?
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okay nice
?tag helping
**When helping people, please do not dump the full solution. **
Remember that these people are trying to learn concepts, and our goal should be to reinforce them. Giving them a solution will not help.
**Instead, give them hints and guide them toward the correct solution. **
You can ask things like "what part did you get stuck on?" or "what have you tried so far?" or give them a small starting hint. This ensures better learning. For more tips, reach out to Schlaumau (@schlaumau).
Did u get 1/5 as final ans
Yes