Angle problem

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FlyingDragon21

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FlyingDragon21

#1 h Mar 18, 2025, 11:36 PM

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In Isosceles triangle ABC where AB equals AC and point D lies on line AB, line CD splits line AB so that AD equals BC. If angle BAC is 20 degrees, what is the measure of angle DCA?

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FlyingDragon21

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FlyingDragon21

#2 h Mar 18, 2025, 11:50 PM

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can anyone solve this? I can't figure it out.

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yegorovm

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yegorovm

#3 h Mar 18, 2025, 11:53 PM

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Could you provide a diagram/picture?

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FlyingDragon21

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FlyingDragon21

#4 h Mar 19, 2025, 12:13 AM

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sure. heres a nice handdrawn picture:

Attachments:

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kamuii

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kamuii

#5 h Mar 19, 2025, 12:24 AM

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i dont think you have enough information for this could be wrong tho

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yegorovm

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yegorovm

#6 h Mar 19, 2025, 12:31 AM

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The drawing is fire.

This post has been edited 1 time. Last edited by yegorovm, Mar 19, 2025, 12:32 AM
Reason: idk

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yegorovm

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yegorovm

#8 h Mar 19, 2025, 12:36 AM

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The problem doesn't make sense. It says that CD splits AB so that AD equals BC but that doesn't make sense.

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Reason: idk

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yegorovm

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yegorovm

#9 h Mar 19, 2025, 12:37 AM

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yegorovm wrote:

The problem doesn't make sense. It says that CD splits AB so that AD equals BC but that doesn't make sense.

Is this just me?

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iwastedmyusername

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iwastedmyusername

#10 h Mar 19, 2025, 12:38 AM

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d is a point on ab so that ad = bc?
anyway I have no clue how to do it

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Reason: aesaseffase

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Shadow6885

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Shadow6885

#11 h Mar 19, 2025, 12:39 AM

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I think this can be solved with trig?
Idk tbh

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EaZ_Shadow

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EaZ_Shadow

#12 h Mar 19, 2025, 12:43 AM

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Bro not enough information

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scannose

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scannose

#13 h Mar 19, 2025, 12:58 AM • 1 Y

Y by ibmo0907

the information provided by the problem statement is sufficient because there is an unique triangle you can construct given the conditions

using law of sines, you have
sin x / sin 20 = ad / dc = bc / dc = sin (20 + x) / sin 80
sin x * sin 80 = sin (20 + x) * sin 20
trial and error gives that x = 10 degrees, as sin 10 * sin 80 = sin 10 * cos 10 = 1/2 * sin 20 = sin 30 * sin 20

i believe there's a more troll solution to this problem, though i forgot what it is

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FlyingDragon21

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FlyingDragon21

#14 h Mar 19, 2025, 1:03 AM

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scannose wrote:

the information provided by the problem statement is sufficient because there is an unique triangle you can construct given the conditions

using law of sines, you have
sin x / sin 20 = ad / dc = bc / dc = sin (20 + x) / sin 80
sin x * sin 80 = sin (20 + x) * sin 20
trial and error gives that x = 10 degrees, as sin 10 * sin 80 = sin 10 * cos 10 = 1/2 * sin 20 = sin 30 * sin 20

i believe there's a more troll solution to this problem, though i forgot what it is

I think you're probably right. I haven't learned sine and cosine yet tho. A friend gave me this question to solve, but he can't solve it either so he asked me. I don't know where he got this question from.

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Alex_Yang

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Alex_Yang

#15 h Mar 19, 2025, 3:37 PM

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FlyingDragon21 wrote:

sure. heres a nice handdrawn picture:

fairly a trivial problem by law of sines:

LOS gives us that $\sin{x}/AD$ is equal to $\sin{20}/DC$ , while $\sin{ABC}/DC$ = $\sin{BDC}/BC$

since AC=BC and BDC= $20+x$ and $ABC$ is $80$

we get that $\sin{x}/AD=\sin{20}/DC$ and $\sin{80}/DC=\sin{(20+x)}/BC$

thus $DC/BC=\sin{80}/\sin{20+x}$ and $DC/AD=\sin{20}/\sin{x}$

since $BC=AD$ we equate the two to get $\sin{80}/\sin{20+x}=\sin{20}/\sin{x}$

just solve for it to get x=10 degrees and we are done

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fruitmonster97

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fruitmonster97

#16 h Mar 19, 2025, 4:46 PM • 1 Y

Y by scannose

this is a very silly problem.

Consider $E$ such that $E$ and $B$ are on opposite sides of $AC$ and $AED\cong BAC.$ Then, we have $\angle CAE=60^\circ$ and $AC=AE,$ so $ACE$ is equilateral. Thus, $AE=DE=CE,$ so $(ADC)$ is centered at $E.$ Thus, $\angle ADC=180^\circ-\tfrac{\angle AEC}{2}=150^\circ,$ so $\angle ACD=180^\circ-150^\circ-20^\circ=\boxed{10^\circ}.$

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Bnn81351

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Bnn81351

#17 h Mar 19, 2025, 11:43 PM

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Given isosceles $\triangle ABC$ with $AB = AC$ and $\angle BAC = 20^\circ$ , we find $\angle ABC = \angle ACB = 80^\circ$ . Since $AD = BC$ , $\triangle ACD$ and $\triangle BCD$ are congruent by SAS. This implies $\angle DCA = \angle BCA = 80^\circ$ . Hence, $\angle DCA = \boxed{80^\circ}$ .

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scannose

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scannose

#18 h Mar 19, 2025, 11:47 PM

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fruitmonster97 wrote:

this is a very silly problem.

Consider $E$ such that $E$ and $B$ are on opposite sides of $AC$ and $AED\cong BAC.$ Then, we have $\angle CAE=60^\circ$ and $AC=AE,$ so $ACE$ is equilateral. Thus, $AE=DE=CE,$ so $(ADC)$ is centered at $E.$ Thus, $\angle ADC=180^\circ-\tfrac{\angle AEC}{2}=150^\circ,$ so $\angle ACD=180^\circ-150^\circ-20^\circ=\boxed{10^\circ}.$

based ty

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FlyingDragon21

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FlyingDragon21

#19 h Mar 20, 2025, 12:12 AM

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fruitmonster97 wrote:

this is a very silly problem.

Consider $E$ such that $E$ and $B$ are on opposite sides of $AC$ and $AED\cong BAC.$ Then, we have $\angle CAE=60^\circ$ and $AC=AE,$ so $ACE$ is equilateral. Thus, $AE=DE=CE,$ so $(ADC)$ is centered at $E.$ Thus, $\angle ADC=180^\circ-\tfrac{\angle AEC}{2}=150^\circ,$ so $\angle ACD=180^\circ-150^\circ-20^\circ=\boxed{10^\circ}.$

wait, what? I don't get why $\angle ADC=180^\circ-\tfrac{\angle AEC}{2}^\circ$ . could someone please explain it?

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Bnn81351

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Bnn81351

#20 h Mar 20, 2025, 12:29 AM

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Bnn81351 wrote:

Given isosceles $\triangle ABC$ with $AB = AC$ and $\angle BAC = 20^\circ$ , we find $\angle ABC = \angle ACB = 80^\circ$ . Since $AD = BC$ , $\triangle ACD$ and $\triangle BCD$ are congruent by SAS. This implies $\angle DCA = \angle BCA = 80^\circ$ . Hence, $\angle DCA = \boxed{80^\circ}$ .

Does this work?

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FlyingDragon21

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FlyingDragon21

#21 h Mar 20, 2025, 12:37 AM

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Bnn81351 wrote:

Given isosceles $\triangle ABC$ with $AB = AC$ and $\angle BAC = 20^\circ$ , we find $\angle ABC = \angle ACB = 80^\circ$ . Since $AD = BC$ , $\triangle ACD$ and $\triangle BCD$ are congruent by SAS. This implies $\angle DCA = \angle BCA = 80^\circ$ . Hence, $\angle DCA = \boxed{80^\circ}$ .

Does this work?

no. your solution suggests that AB = CB, which is not true.

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Bnn81351

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Bnn81351

#22 h Mar 20, 2025, 12:38 AM

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oh okay

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FlyingDragon21

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FlyingDragon21

#23 h Mar 20, 2025, 12:40 AM

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FlyingDragon21 wrote:

fruitmonster97 wrote:

this is a very silly problem.

Consider $E$ such that $E$ and $B$ are on opposite sides of $AC$ and $AED\cong BAC.$ Then, we have $\angle CAE=60^\circ$ and $AC=AE,$ so $ACE$ is equilateral. Thus, $AE=DE=CE,$ so $(ADC)$ is centered at $E.$ Thus, $\angle ADC=180^\circ-\tfrac{\angle AEC}{2}=150^\circ,$ so $\angle ACD=180^\circ-150^\circ-20^\circ=\boxed{10^\circ}.$

wait, what? I don't get why $\angle ADC=180^\circ-\tfrac{\angle AEC}{2}^\circ$ . could someone please explain it?

@Bnn81351, do you understand this though?

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Bnn81351

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Bnn81351

#24 h Mar 20, 2025, 12:40 AM

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yeah yeah yeahI know

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fruitmonster97

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fruitmonster97

#25 h Thursday at 12:17 PM • 1 Y

Y by FlyingDragon21

FlyingDragon21 wrote:

wait, what? I don't get why $\angle ADC=180^\circ-\tfrac{\angle AEC}{2}^\circ$ . could someone please explain it?

pick $P$ on major arc $CD$ of $(E).$ Then by cyclic quad $\angle ADC=180^\circ-\angle APC,$ and by inscribed angle theorem $\angle APC=\tfrac{\angle AEC}{2}.$ Thus, you get $\angle ADC=180^\circ-\tfrac{\angle AEC}{2}.$

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FlyingDragon21

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FlyingDragon21

#26 h Thursday at 8:16 PM

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fruitmonster97 wrote:

FlyingDragon21 wrote:

wait, what? I don't get why $\angle ADC=180^\circ-\tfrac{\angle AEC}{2}^\circ$ . could someone please explain it?

pick $P$ on major arc $CD$ of $(E).$ Then by cyclic quad $\angle ADC=180^\circ-\angle APC,$ and by inscribed angle theorem $\angle APC=\tfrac{\angle AEC}{2}.$ Thus, you get $\angle ADC=180^\circ-\tfrac{\angle AEC}{2}.$

oh, ok. I get it. Thanks!

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