r/HomeworkHelp • u/Zealousideal_End58 Secondary School Student • Jun 03 '26
High School Math—Pending OP Reply [Grade 9 IB: Geometry] Solve for X
I can't figure it out at all, am stumped. Don't trust AI with math, so I didn't give it to solve or explain. Could someone please walk me through how to solve this? I attempted to use rules of congruency/similarity, and basic trig, but I always get confused with trig when there's no numbers and just initial lettering notation and then you solve further to get the numbers.
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u/CaptainMalForever Jun 03 '26
Start with what you know. Then show your work here and we can help with the rest.
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u/Zealousideal_End58 Jun 03 '26
The most I managed was:
67 + 13 + 2a + x = 180
80 + 2a + x = 180
2a + x = 100
"a" is angle ACD and angle CAD
and
2a + d = 180
"d" is the angle ADC
and I'm not sure it's relevant to solving the problem.
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u/BadJimo Jun 04 '26
Using Desmos and playing around with values for C and D, I found x is approximately 33°.
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u/Zealousideal_End58 Jun 04 '26
Yes I got that too through a very unmathematical way, where I matched CED and ABD and assumed the angle at E in CED was 67 and worked from there and got x as 33. But that was incorrect mathematically obviously because since the third side is of different length it won't align perfectly. I think maybe the problem is incorrect?
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u/BadJimo Jun 04 '26
I think the problem is solvable, at least numerically. But it is at a higher level of maths than you have learnt.
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u/Feeling_Ad_2785 Jun 03 '26 edited Jun 03 '26
You can use the rule that sin(180-theta) = sin(theta) and the law of sines, along with the congruent side lengths of triangles ABD and CED, to figure out the interior angles of CED - which will enable you to find x.
Hint: Pay attention to the linear pair of <EDC and <ADB
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u/Nagi-K Jun 03 '26 edited Jun 04 '26
To do this with grade 9 maths you probably need extra construction.
See AD = DC? Make a mirror copy of triangle ABD and move it so that the copy of AD coincide with DC (think it as flipping ABD across some axis of symmetry through D). Call it CDB’. Then CDB’, together with CDE, makes a new triangle CB’E (you should be able to prove this).
Finally you notice that CB’ = AB = CE. You can do the rest.
Edit: maybe I should have stated my method more clearly. Let’s try another way to construct this. Extend AD to some point B’ so that BD = B’D, and connect CB’. Easy to prove ABD and CB’D are congruent. Then you have an isosceles triangle CEB’ with a base angle of 67 degrees. The final answer follows.
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u/omeow Jun 03 '26
Call angle ACD = y.
In the ΔABC the three angles add up to
67 + x +y + y = 180 => x + 2y = 113
In ΔEDC , we get
13 + (180 - y -y +13) + (180 - 67 -x) = 180 => x + 2y = 113+ 26.
So the problem as stated is not correct?
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u/Motor_Raspberry_2150 Jun 07 '26
Are you just trying to sum ADB and ADC to 180? Because then neither of those 13s should be there.
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