Two equal segments, two known angles, find angle in triangle












2












$begingroup$


enter image description here



This is the problem. I have tried the obvious without success. Drawing some perpendiculars (from A for example), I only know the angle $sphericalangle ADC = 54 ^{circ}$. How can I use the equality $AB = DC$?






geometry trigonometry euclidean-geometry geometric-transformation







share|cite|improve this question











$endgroup$












  • $begingroup$
    A small point is that $angle ADC = 55°$, not $54°$ as you state, based on $angle ABC = 25°$ and $angle BAD = 30°$. Also, it's not clear to me whether it's $DC$ or $BC$ which is equal in length to $AB$. Please clarify this. Thanks. Note that if it's $BC$, then you have an isosceles triangle, so it's quite simple that $x$ is $155/2°$, but then having $angle BAD = 30°$ doesn't really do anything.
    $endgroup$
    – John Omielan
    18 hours ago












  • $begingroup$
    Thx for the comment. AB=DC. And the angle ABC is 24, not 25.
    $endgroup$
    – asd11
    17 hours ago










  • $begingroup$
    You are welcome. The writing made it seem like it was $25$ instead to me.
    $endgroup$
    – John Omielan
    17 hours ago
















2












$begingroup$


enter image description here



This is the problem. I have tried the obvious without success. Drawing some perpendiculars (from A for example), I only know the angle $sphericalangle ADC = 54 ^{circ}$. How can I use the equality $AB = DC$?






geometry trigonometry euclidean-geometry geometric-transformation







share|cite|improve this question











$endgroup$












  • $begingroup$
    A small point is that $angle ADC = 55°$, not $54°$ as you state, based on $angle ABC = 25°$ and $angle BAD = 30°$. Also, it's not clear to me whether it's $DC$ or $BC$ which is equal in length to $AB$. Please clarify this. Thanks. Note that if it's $BC$, then you have an isosceles triangle, so it's quite simple that $x$ is $155/2°$, but then having $angle BAD = 30°$ doesn't really do anything.
    $endgroup$
    – John Omielan
    18 hours ago












  • $begingroup$
    Thx for the comment. AB=DC. And the angle ABC is 24, not 25.
    $endgroup$
    – asd11
    17 hours ago










  • $begingroup$
    You are welcome. The writing made it seem like it was $25$ instead to me.
    $endgroup$
    – John Omielan
    17 hours ago














2












2








2


1



$begingroup$


enter image description here



This is the problem. I have tried the obvious without success. Drawing some perpendiculars (from A for example), I only know the angle $sphericalangle ADC = 54 ^{circ}$. How can I use the equality $AB = DC$?






geometry trigonometry euclidean-geometry geometric-transformation







share|cite|improve this question











$endgroup$




enter image description here



This is the problem. I have tried the obvious without success. Drawing some perpendiculars (from A for example), I only know the angle $sphericalangle ADC = 54 ^{circ}$. How can I use the equality $AB = DC$?






geometry trigonometry euclidean-geometry geometric-transformation




geometry trigonometry euclidean-geometry geometric-transformation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 11 hours ago









greedoid

42k1152105




42k1152105










asked 18 hours ago









asd11asd11

30518




30518












  • $begingroup$
    A small point is that $angle ADC = 55°$, not $54°$ as you state, based on $angle ABC = 25°$ and $angle BAD = 30°$. Also, it's not clear to me whether it's $DC$ or $BC$ which is equal in length to $AB$. Please clarify this. Thanks. Note that if it's $BC$, then you have an isosceles triangle, so it's quite simple that $x$ is $155/2°$, but then having $angle BAD = 30°$ doesn't really do anything.
    $endgroup$
    – John Omielan
    18 hours ago












  • $begingroup$
    Thx for the comment. AB=DC. And the angle ABC is 24, not 25.
    $endgroup$
    – asd11
    17 hours ago










  • $begingroup$
    You are welcome. The writing made it seem like it was $25$ instead to me.
    $endgroup$
    – John Omielan
    17 hours ago


















  • $begingroup$
    A small point is that $angle ADC = 55°$, not $54°$ as you state, based on $angle ABC = 25°$ and $angle BAD = 30°$. Also, it's not clear to me whether it's $DC$ or $BC$ which is equal in length to $AB$. Please clarify this. Thanks. Note that if it's $BC$, then you have an isosceles triangle, so it's quite simple that $x$ is $155/2°$, but then having $angle BAD = 30°$ doesn't really do anything.
    $endgroup$
    – John Omielan
    18 hours ago












  • $begingroup$
    Thx for the comment. AB=DC. And the angle ABC is 24, not 25.
    $endgroup$
    – asd11
    17 hours ago










  • $begingroup$
    You are welcome. The writing made it seem like it was $25$ instead to me.
    $endgroup$
    – John Omielan
    17 hours ago
















$begingroup$
A small point is that $angle ADC = 55°$, not $54°$ as you state, based on $angle ABC = 25°$ and $angle BAD = 30°$. Also, it's not clear to me whether it's $DC$ or $BC$ which is equal in length to $AB$. Please clarify this. Thanks. Note that if it's $BC$, then you have an isosceles triangle, so it's quite simple that $x$ is $155/2°$, but then having $angle BAD = 30°$ doesn't really do anything.
$endgroup$
– John Omielan
18 hours ago






$begingroup$
A small point is that $angle ADC = 55°$, not $54°$ as you state, based on $angle ABC = 25°$ and $angle BAD = 30°$. Also, it's not clear to me whether it's $DC$ or $BC$ which is equal in length to $AB$. Please clarify this. Thanks. Note that if it's $BC$, then you have an isosceles triangle, so it's quite simple that $x$ is $155/2°$, but then having $angle BAD = 30°$ doesn't really do anything.
$endgroup$
– John Omielan
18 hours ago














$begingroup$
Thx for the comment. AB=DC. And the angle ABC is 24, not 25.
$endgroup$
– asd11
17 hours ago




$begingroup$
Thx for the comment. AB=DC. And the angle ABC is 24, not 25.
$endgroup$
– asd11
17 hours ago












$begingroup$
You are welcome. The writing made it seem like it was $25$ instead to me.
$endgroup$
– John Omielan
17 hours ago




$begingroup$
You are welcome. The writing made it seem like it was $25$ instead to me.
$endgroup$
– John Omielan
17 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

By the law of sines we obtain:
$$frac{AD}{sin{x}}=frac{DC}{sin(126^{circ}-x)}$$ and
$$frac{AD}{sin24^{circ}}=frac{AB}{sin54^{circ}}$$ and since $AB=DC$, we obtain:
$$frac{sin{x}}{sin24^{circ}}=frac{sin(126^{circ}-x)}{sin54^{circ}}.$$
Can you end it now?



I got $x=30^{circ}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If you do that trig, then you'll find that BD:DC is a ratio the OP might recognise. Wonder if there's a way to prove it geometrically. (From such a proof, @Michael's result would follow easily.)
    $endgroup$
    – Rosie F
    16 hours ago










  • $begingroup$
    @Rosie F I still don't see a geometric solution.
    $endgroup$
    – Michael Rozenberg
    16 hours ago










  • $begingroup$
    I really wonder how much are calculations here faster than pure geometric solution?
    $endgroup$
    – greedoid
    13 hours ago



















3












$begingroup$

Let's reformulate the problem.




Let $triangle ABC$ be such that $angle ABC = 24^{circ}$ and $angle ACB = 30^{circ}$ and $D$ is on $BC$ so that $AB = CD$.



We have to prove $angle BAD = 30^{circ}$




enter image description here



Solution: Rotate $B$ around $A$ for $60^{circ}$ to new point $E$. Then since $EA=EB$ and $$angle AEB = 2angle BCA$$ we see that $C$ is on circle with center at $E$ and radius $EA=EB = EC$. Then $CD = CE$, so $$angle BCE = angle CBE = 34^{circ} implies angle EDC = angle DEC = 72^{circ}$$
Now this means that $$angle BDE = 108^{circ} implies angle BED = angle EBD = 36^{circ}$$
This means $BD = ED$, so so $triangle ABD$ and $triangle ADE$ are congruent (sss) and thus $angle BAD = 30 ^{circ}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Very nice solution! +1
    $endgroup$
    – Michael Rozenberg
    13 hours ago










  • $begingroup$
    Ah, glad that this angle problem with its rational answer has a geometrical solution. Classical elements of an adventitious-angle problem's solution: an equilateral triangle and a circumcentre. Pity it's analytical (with your reformulation) but it's sound nonetheless.
    $endgroup$
    – Rosie F
    5 hours ago










  • $begingroup$
    Which part in the solution is analytical? @RosieF
    $endgroup$
    – greedoid
    5 hours ago












  • $begingroup$
    @greedoid You needed to know that the answer is $30^circ$ in order to know to make $angle ACB$ $30^circ$ in your reformulation. I acknowledge that your solution is sound -- seeing as making $angle ACB$ $30^circ$ implies $angle BAD=30^circ$, the other constraints mean that, conversely, if $angle BAD=30^circ$ as specified, $angle ACB=30^circ$. But your reformulation is analytical in that it analysed the answer and proved it correct rather than synthesizing it directly.
    $endgroup$
    – Rosie F
    5 hours ago












  • $begingroup$
    I'm not sure what you mean an but easy drawing make easy hypothethis it is 30 degree.@RosieF
    $endgroup$
    – greedoid
    5 hours ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

By the law of sines we obtain:
$$frac{AD}{sin{x}}=frac{DC}{sin(126^{circ}-x)}$$ and
$$frac{AD}{sin24^{circ}}=frac{AB}{sin54^{circ}}$$ and since $AB=DC$, we obtain:
$$frac{sin{x}}{sin24^{circ}}=frac{sin(126^{circ}-x)}{sin54^{circ}}.$$
Can you end it now?



I got $x=30^{circ}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If you do that trig, then you'll find that BD:DC is a ratio the OP might recognise. Wonder if there's a way to prove it geometrically. (From such a proof, @Michael's result would follow easily.)
    $endgroup$
    – Rosie F
    16 hours ago










  • $begingroup$
    @Rosie F I still don't see a geometric solution.
    $endgroup$
    – Michael Rozenberg
    16 hours ago










  • $begingroup$
    I really wonder how much are calculations here faster than pure geometric solution?
    $endgroup$
    – greedoid
    13 hours ago
















4












$begingroup$

By the law of sines we obtain:
$$frac{AD}{sin{x}}=frac{DC}{sin(126^{circ}-x)}$$ and
$$frac{AD}{sin24^{circ}}=frac{AB}{sin54^{circ}}$$ and since $AB=DC$, we obtain:
$$frac{sin{x}}{sin24^{circ}}=frac{sin(126^{circ}-x)}{sin54^{circ}}.$$
Can you end it now?



I got $x=30^{circ}.$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If you do that trig, then you'll find that BD:DC is a ratio the OP might recognise. Wonder if there's a way to prove it geometrically. (From such a proof, @Michael's result would follow easily.)
    $endgroup$
    – Rosie F
    16 hours ago










  • $begingroup$
    @Rosie F I still don't see a geometric solution.
    $endgroup$
    – Michael Rozenberg
    16 hours ago










  • $begingroup$
    I really wonder how much are calculations here faster than pure geometric solution?
    $endgroup$
    – greedoid
    13 hours ago














4












4








4





$begingroup$

By the law of sines we obtain:
$$frac{AD}{sin{x}}=frac{DC}{sin(126^{circ}-x)}$$ and
$$frac{AD}{sin24^{circ}}=frac{AB}{sin54^{circ}}$$ and since $AB=DC$, we obtain:
$$frac{sin{x}}{sin24^{circ}}=frac{sin(126^{circ}-x)}{sin54^{circ}}.$$
Can you end it now?



I got $x=30^{circ}.$






share|cite|improve this answer









$endgroup$



By the law of sines we obtain:
$$frac{AD}{sin{x}}=frac{DC}{sin(126^{circ}-x)}$$ and
$$frac{AD}{sin24^{circ}}=frac{AB}{sin54^{circ}}$$ and since $AB=DC$, we obtain:
$$frac{sin{x}}{sin24^{circ}}=frac{sin(126^{circ}-x)}{sin54^{circ}}.$$
Can you end it now?



I got $x=30^{circ}.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 17 hours ago









Michael RozenbergMichael Rozenberg

103k1891195




103k1891195












  • $begingroup$
    If you do that trig, then you'll find that BD:DC is a ratio the OP might recognise. Wonder if there's a way to prove it geometrically. (From such a proof, @Michael's result would follow easily.)
    $endgroup$
    – Rosie F
    16 hours ago










  • $begingroup$
    @Rosie F I still don't see a geometric solution.
    $endgroup$
    – Michael Rozenberg
    16 hours ago










  • $begingroup$
    I really wonder how much are calculations here faster than pure geometric solution?
    $endgroup$
    – greedoid
    13 hours ago


















  • $begingroup$
    If you do that trig, then you'll find that BD:DC is a ratio the OP might recognise. Wonder if there's a way to prove it geometrically. (From such a proof, @Michael's result would follow easily.)
    $endgroup$
    – Rosie F
    16 hours ago










  • $begingroup$
    @Rosie F I still don't see a geometric solution.
    $endgroup$
    – Michael Rozenberg
    16 hours ago










  • $begingroup$
    I really wonder how much are calculations here faster than pure geometric solution?
    $endgroup$
    – greedoid
    13 hours ago
















$begingroup$
If you do that trig, then you'll find that BD:DC is a ratio the OP might recognise. Wonder if there's a way to prove it geometrically. (From such a proof, @Michael's result would follow easily.)
$endgroup$
– Rosie F
16 hours ago




$begingroup$
If you do that trig, then you'll find that BD:DC is a ratio the OP might recognise. Wonder if there's a way to prove it geometrically. (From such a proof, @Michael's result would follow easily.)
$endgroup$
– Rosie F
16 hours ago












$begingroup$
@Rosie F I still don't see a geometric solution.
$endgroup$
– Michael Rozenberg
16 hours ago




$begingroup$
@Rosie F I still don't see a geometric solution.
$endgroup$
– Michael Rozenberg
16 hours ago












$begingroup$
I really wonder how much are calculations here faster than pure geometric solution?
$endgroup$
– greedoid
13 hours ago




$begingroup$
I really wonder how much are calculations here faster than pure geometric solution?
$endgroup$
– greedoid
13 hours ago











3












$begingroup$

Let's reformulate the problem.




Let $triangle ABC$ be such that $angle ABC = 24^{circ}$ and $angle ACB = 30^{circ}$ and $D$ is on $BC$ so that $AB = CD$.



We have to prove $angle BAD = 30^{circ}$




enter image description here



Solution: Rotate $B$ around $A$ for $60^{circ}$ to new point $E$. Then since $EA=EB$ and $$angle AEB = 2angle BCA$$ we see that $C$ is on circle with center at $E$ and radius $EA=EB = EC$. Then $CD = CE$, so $$angle BCE = angle CBE = 34^{circ} implies angle EDC = angle DEC = 72^{circ}$$
Now this means that $$angle BDE = 108^{circ} implies angle BED = angle EBD = 36^{circ}$$
This means $BD = ED$, so so $triangle ABD$ and $triangle ADE$ are congruent (sss) and thus $angle BAD = 30 ^{circ}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Very nice solution! +1
    $endgroup$
    – Michael Rozenberg
    13 hours ago










  • $begingroup$
    Ah, glad that this angle problem with its rational answer has a geometrical solution. Classical elements of an adventitious-angle problem's solution: an equilateral triangle and a circumcentre. Pity it's analytical (with your reformulation) but it's sound nonetheless.
    $endgroup$
    – Rosie F
    5 hours ago










  • $begingroup$
    Which part in the solution is analytical? @RosieF
    $endgroup$
    – greedoid
    5 hours ago












  • $begingroup$
    @greedoid You needed to know that the answer is $30^circ$ in order to know to make $angle ACB$ $30^circ$ in your reformulation. I acknowledge that your solution is sound -- seeing as making $angle ACB$ $30^circ$ implies $angle BAD=30^circ$, the other constraints mean that, conversely, if $angle BAD=30^circ$ as specified, $angle ACB=30^circ$. But your reformulation is analytical in that it analysed the answer and proved it correct rather than synthesizing it directly.
    $endgroup$
    – Rosie F
    5 hours ago












  • $begingroup$
    I'm not sure what you mean an but easy drawing make easy hypothethis it is 30 degree.@RosieF
    $endgroup$
    – greedoid
    5 hours ago


















3












$begingroup$

Let's reformulate the problem.




Let $triangle ABC$ be such that $angle ABC = 24^{circ}$ and $angle ACB = 30^{circ}$ and $D$ is on $BC$ so that $AB = CD$.



We have to prove $angle BAD = 30^{circ}$




enter image description here



Solution: Rotate $B$ around $A$ for $60^{circ}$ to new point $E$. Then since $EA=EB$ and $$angle AEB = 2angle BCA$$ we see that $C$ is on circle with center at $E$ and radius $EA=EB = EC$. Then $CD = CE$, so $$angle BCE = angle CBE = 34^{circ} implies angle EDC = angle DEC = 72^{circ}$$
Now this means that $$angle BDE = 108^{circ} implies angle BED = angle EBD = 36^{circ}$$
This means $BD = ED$, so so $triangle ABD$ and $triangle ADE$ are congruent (sss) and thus $angle BAD = 30 ^{circ}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Very nice solution! +1
    $endgroup$
    – Michael Rozenberg
    13 hours ago










  • $begingroup$
    Ah, glad that this angle problem with its rational answer has a geometrical solution. Classical elements of an adventitious-angle problem's solution: an equilateral triangle and a circumcentre. Pity it's analytical (with your reformulation) but it's sound nonetheless.
    $endgroup$
    – Rosie F
    5 hours ago










  • $begingroup$
    Which part in the solution is analytical? @RosieF
    $endgroup$
    – greedoid
    5 hours ago












  • $begingroup$
    @greedoid You needed to know that the answer is $30^circ$ in order to know to make $angle ACB$ $30^circ$ in your reformulation. I acknowledge that your solution is sound -- seeing as making $angle ACB$ $30^circ$ implies $angle BAD=30^circ$, the other constraints mean that, conversely, if $angle BAD=30^circ$ as specified, $angle ACB=30^circ$. But your reformulation is analytical in that it analysed the answer and proved it correct rather than synthesizing it directly.
    $endgroup$
    – Rosie F
    5 hours ago












  • $begingroup$
    I'm not sure what you mean an but easy drawing make easy hypothethis it is 30 degree.@RosieF
    $endgroup$
    – greedoid
    5 hours ago
















3












3








3





$begingroup$

Let's reformulate the problem.




Let $triangle ABC$ be such that $angle ABC = 24^{circ}$ and $angle ACB = 30^{circ}$ and $D$ is on $BC$ so that $AB = CD$.



We have to prove $angle BAD = 30^{circ}$




enter image description here



Solution: Rotate $B$ around $A$ for $60^{circ}$ to new point $E$. Then since $EA=EB$ and $$angle AEB = 2angle BCA$$ we see that $C$ is on circle with center at $E$ and radius $EA=EB = EC$. Then $CD = CE$, so $$angle BCE = angle CBE = 34^{circ} implies angle EDC = angle DEC = 72^{circ}$$
Now this means that $$angle BDE = 108^{circ} implies angle BED = angle EBD = 36^{circ}$$
This means $BD = ED$, so so $triangle ABD$ and $triangle ADE$ are congruent (sss) and thus $angle BAD = 30 ^{circ}$.






share|cite|improve this answer











$endgroup$



Let's reformulate the problem.




Let $triangle ABC$ be such that $angle ABC = 24^{circ}$ and $angle ACB = 30^{circ}$ and $D$ is on $BC$ so that $AB = CD$.



We have to prove $angle BAD = 30^{circ}$




enter image description here



Solution: Rotate $B$ around $A$ for $60^{circ}$ to new point $E$. Then since $EA=EB$ and $$angle AEB = 2angle BCA$$ we see that $C$ is on circle with center at $E$ and radius $EA=EB = EC$. Then $CD = CE$, so $$angle BCE = angle CBE = 34^{circ} implies angle EDC = angle DEC = 72^{circ}$$
Now this means that $$angle BDE = 108^{circ} implies angle BED = angle EBD = 36^{circ}$$
This means $BD = ED$, so so $triangle ABD$ and $triangle ADE$ are congruent (sss) and thus $angle BAD = 30 ^{circ}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 13 hours ago

























answered 14 hours ago









greedoidgreedoid

42k1152105




42k1152105












  • $begingroup$
    Very nice solution! +1
    $endgroup$
    – Michael Rozenberg
    13 hours ago










  • $begingroup$
    Ah, glad that this angle problem with its rational answer has a geometrical solution. Classical elements of an adventitious-angle problem's solution: an equilateral triangle and a circumcentre. Pity it's analytical (with your reformulation) but it's sound nonetheless.
    $endgroup$
    – Rosie F
    5 hours ago










  • $begingroup$
    Which part in the solution is analytical? @RosieF
    $endgroup$
    – greedoid
    5 hours ago












  • $begingroup$
    @greedoid You needed to know that the answer is $30^circ$ in order to know to make $angle ACB$ $30^circ$ in your reformulation. I acknowledge that your solution is sound -- seeing as making $angle ACB$ $30^circ$ implies $angle BAD=30^circ$, the other constraints mean that, conversely, if $angle BAD=30^circ$ as specified, $angle ACB=30^circ$. But your reformulation is analytical in that it analysed the answer and proved it correct rather than synthesizing it directly.
    $endgroup$
    – Rosie F
    5 hours ago












  • $begingroup$
    I'm not sure what you mean an but easy drawing make easy hypothethis it is 30 degree.@RosieF
    $endgroup$
    – greedoid
    5 hours ago




















  • $begingroup$
    Very nice solution! +1
    $endgroup$
    – Michael Rozenberg
    13 hours ago










  • $begingroup$
    Ah, glad that this angle problem with its rational answer has a geometrical solution. Classical elements of an adventitious-angle problem's solution: an equilateral triangle and a circumcentre. Pity it's analytical (with your reformulation) but it's sound nonetheless.
    $endgroup$
    – Rosie F
    5 hours ago










  • $begingroup$
    Which part in the solution is analytical? @RosieF
    $endgroup$
    – greedoid
    5 hours ago












  • $begingroup$
    @greedoid You needed to know that the answer is $30^circ$ in order to know to make $angle ACB$ $30^circ$ in your reformulation. I acknowledge that your solution is sound -- seeing as making $angle ACB$ $30^circ$ implies $angle BAD=30^circ$, the other constraints mean that, conversely, if $angle BAD=30^circ$ as specified, $angle ACB=30^circ$. But your reformulation is analytical in that it analysed the answer and proved it correct rather than synthesizing it directly.
    $endgroup$
    – Rosie F
    5 hours ago












  • $begingroup$
    I'm not sure what you mean an but easy drawing make easy hypothethis it is 30 degree.@RosieF
    $endgroup$
    – greedoid
    5 hours ago


















$begingroup$
Very nice solution! +1
$endgroup$
– Michael Rozenberg
13 hours ago




$begingroup$
Very nice solution! +1
$endgroup$
– Michael Rozenberg
13 hours ago












$begingroup$
Ah, glad that this angle problem with its rational answer has a geometrical solution. Classical elements of an adventitious-angle problem's solution: an equilateral triangle and a circumcentre. Pity it's analytical (with your reformulation) but it's sound nonetheless.
$endgroup$
– Rosie F
5 hours ago




$begingroup$
Ah, glad that this angle problem with its rational answer has a geometrical solution. Classical elements of an adventitious-angle problem's solution: an equilateral triangle and a circumcentre. Pity it's analytical (with your reformulation) but it's sound nonetheless.
$endgroup$
– Rosie F
5 hours ago












$begingroup$
Which part in the solution is analytical? @RosieF
$endgroup$
– greedoid
5 hours ago






$begingroup$
Which part in the solution is analytical? @RosieF
$endgroup$
– greedoid
5 hours ago














$begingroup$
@greedoid You needed to know that the answer is $30^circ$ in order to know to make $angle ACB$ $30^circ$ in your reformulation. I acknowledge that your solution is sound -- seeing as making $angle ACB$ $30^circ$ implies $angle BAD=30^circ$, the other constraints mean that, conversely, if $angle BAD=30^circ$ as specified, $angle ACB=30^circ$. But your reformulation is analytical in that it analysed the answer and proved it correct rather than synthesizing it directly.
$endgroup$
– Rosie F
5 hours ago






$begingroup$
@greedoid You needed to know that the answer is $30^circ$ in order to know to make $angle ACB$ $30^circ$ in your reformulation. I acknowledge that your solution is sound -- seeing as making $angle ACB$ $30^circ$ implies $angle BAD=30^circ$, the other constraints mean that, conversely, if $angle BAD=30^circ$ as specified, $angle ACB=30^circ$. But your reformulation is analytical in that it analysed the answer and proved it correct rather than synthesizing it directly.
$endgroup$
– Rosie F
5 hours ago














$begingroup$
I'm not sure what you mean an but easy drawing make easy hypothethis it is 30 degree.@RosieF
$endgroup$
– greedoid
5 hours ago






$begingroup$
I'm not sure what you mean an but easy drawing make easy hypothethis it is 30 degree.@RosieF
$endgroup$
– greedoid
5 hours ago




















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