Mixing
Angle in a 80-80-20 triangle
$begingroup$
Let $ABC$ be isosceles triangle $(AB=AC)$ with $angle BAC = 20^{circ}$. Point $D$ is on side $AC$ such that $angle DBC = 60^{circ}$. Point $E$ is on side $AB$ such that $angle ECB = 50^{circ}$. Find $angle EDB $
Solution:
Let $x=angle EDB $. We see that $BC = BE$
By sinus theorem for $BCE$ and $BCD$ we have $${sin xover sin (x+20^{circ})}={BEover BD} = {sin 40^{circ}over sin 80^{circ}} = {1over 2cos 40^{circ}} $$
From here we get $$tan x = {sin 20^{circ}over 2cos 40^{circ}-cos 20^{circ}}$$
and it is not difficult to see (after some algebraic manipulation) that $x = 30^{circ}$
Here is a question: Is there a neat synthetic soultion to this problem?
trigonometry euclidean-geometry
asked Apr 9 '18 at 19:49
greedoidgreedoid
38.7k114797
$endgroup$
|
show 4 more comments
$begingroup$
Let $ABC$ be isosceles triangle $(AB=AC)$ with $angle BAC = 20^{circ}$. Point $D$ is on side $AC$ such that $angle DBC = 60^{circ}$. Point $E$ is on side $AB$ such that $angle ECB = 50^{circ}$. Find $angle EDB $
Solution:
Let $x=angle EDB $. We see that $BC = BE$
By sinus theorem for $BCE$ and $BCD$ we have $${sin xover sin (x+20^{circ})}={BEover BD} = {sin 40^{circ}over sin 80^{circ}} = {1over 2cos 40^{circ}} $$
From here we get $$tan x = {sin 20^{circ}over 2cos 40^{circ}-cos 20^{circ}}$$
and it is not difficult to see (after some algebraic manipulation) that $x = 30^{circ}$
Here is a question: Is there a neat synthetic soultion to this problem?
trigonometry euclidean-geometry
asked Apr 9 '18 at 19:49
greedoidgreedoid
38.7k114797
$endgroup$
4
$begingroup$
cut-the-knot.org/triangle/80-80-20/IndexToClassical.shtml
$endgroup$
– Aretino
Apr 9 '18 at 19:52
$begingroup$
OMG, thanks!!!!
$endgroup$
– greedoid
Apr 9 '18 at 19:53
$begingroup$
Hmm, should I delete this one now?
$endgroup$
– greedoid
Apr 9 '18 at 19:54
$begingroup$
@ChristianF, no, Aretino should turn that comment into an answer :)
$endgroup$
– The Chaz 2.0
Apr 9 '18 at 19:55
1
$begingroup$
Once upon a time, this problem let me win a competition :D Here it is another reference: static.elitesecurity.org/uploads/1/9/1987242/…
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 20:01
|
show 4 more comments
$begingroup$
Let $ABC$ be isosceles triangle $(AB=AC)$ with $angle BAC = 20^{circ}$. Point $D$ is on side $AC$ such that $angle DBC = 60^{circ}$. Point $E$ is on side $AB$ such that $angle ECB = 50^{circ}$. Find $angle EDB $
Solution:
Let $x=angle EDB $. We see that $BC = BE$
By sinus theorem for $BCE$ and $BCD$ we have $${sin xover sin (x+20^{circ})}={BEover BD} = {sin 40^{circ}over sin 80^{circ}} = {1over 2cos 40^{circ}} $$
From here we get $$tan x = {sin 20^{circ}over 2cos 40^{circ}-cos 20^{circ}}$$
and it is not difficult to see (after some algebraic manipulation) that $x = 30^{circ}$
Here is a question: Is there a neat synthetic soultion to this problem?
trigonometry euclidean-geometry
asked Apr 9 '18 at 19:49
greedoidgreedoid
38.7k114797
$endgroup$
Let $ABC$ be isosceles triangle $(AB=AC)$ with $angle BAC = 20^{circ}$. Point $D$ is on side $AC$ such that $angle DBC = 60^{circ}$. Point $E$ is on side $AB$ such that $angle ECB = 50^{circ}$. Find $angle EDB $
Solution:
Let $x=angle EDB $. We see that $BC = BE$
By sinus theorem for $BCE$ and $BCD$ we have $${sin xover sin (x+20^{circ})}={BEover BD} = {sin 40^{circ}over sin 80^{circ}} = {1over 2cos 40^{circ}} $$
From here we get $$tan x = {sin 20^{circ}over 2cos 40^{circ}-cos 20^{circ}}$$
and it is not difficult to see (after some algebraic manipulation) that $x = 30^{circ}$
Here is a question: Is there a neat synthetic soultion to this problem?
trigonometry euclidean-geometry
trigonometry euclidean-geometry
asked Apr 9 '18 at 19:49
greedoidgreedoid
38.7k114797
asked Apr 9 '18 at 19:49
greedoidgreedoid
38.7k114797
edited Jan 1 at 22:06
greedoid
asked Apr 9 '18 at 19:49
greedoidgreedoid
38.7k114797
asked Apr 9 '18 at 19:49
greedoidgreedoid
38.7k114797
asked Apr 9 '18 at 19:49
greedoidgreedoid
38.7k114797
38.7k114797
4
$begingroup$
cut-the-knot.org/triangle/80-80-20/IndexToClassical.shtml
$endgroup$
– Aretino
Apr 9 '18 at 19:52
$begingroup$
OMG, thanks!!!!
$endgroup$
– greedoid
Apr 9 '18 at 19:53
$begingroup$
Hmm, should I delete this one now?
$endgroup$
– greedoid
Apr 9 '18 at 19:54
$begingroup$
@ChristianF, no, Aretino should turn that comment into an answer :)
$endgroup$
– The Chaz 2.0
Apr 9 '18 at 19:55
1
$begingroup$
Once upon a time, this problem let me win a competition :D Here it is another reference: static.elitesecurity.org/uploads/1/9/1987242/…
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 20:01
|
show 4 more comments
4
$begingroup$
cut-the-knot.org/triangle/80-80-20/IndexToClassical.shtml
$endgroup$
– Aretino
Apr 9 '18 at 19:52
$begingroup$
OMG, thanks!!!!
$endgroup$
– greedoid
Apr 9 '18 at 19:53
$begingroup$
Hmm, should I delete this one now?
$endgroup$
– greedoid
Apr 9 '18 at 19:54
$begingroup$
@ChristianF, no, Aretino should turn that comment into an answer :)
$endgroup$
– The Chaz 2.0
Apr 9 '18 at 19:55
1
$begingroup$
Once upon a time, this problem let me win a competition :D Here it is another reference: static.elitesecurity.org/uploads/1/9/1987242/…
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 20:01
4
4
$begingroup$
cut-the-knot.org/triangle/80-80-20/IndexToClassical.shtml
$endgroup$
– Aretino
Apr 9 '18 at 19:52
$begingroup$
cut-the-knot.org/triangle/80-80-20/IndexToClassical.shtml
$endgroup$
– Aretino
Apr 9 '18 at 19:52
$begingroup$
OMG, thanks!!!!
$endgroup$
– greedoid
Apr 9 '18 at 19:53
$begingroup$
OMG, thanks!!!!
$endgroup$
– greedoid
Apr 9 '18 at 19:53
$begingroup$
Hmm, should I delete this one now?
$endgroup$
– greedoid
Apr 9 '18 at 19:54
$begingroup$
Hmm, should I delete this one now?
$endgroup$
– greedoid
Apr 9 '18 at 19:54
$begingroup$
@ChristianF, no, Aretino should turn that comment into an answer :)
$endgroup$
– The Chaz 2.0
Apr 9 '18 at 19:55
$begingroup$
@ChristianF, no, Aretino should turn that comment into an answer :)
$endgroup$
– The Chaz 2.0
Apr 9 '18 at 19:55
1
1
$begingroup$
Once upon a time, this problem let me win a competition :D Here it is another reference: static.elitesecurity.org/uploads/1/9/1987242/…
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 20:01
$begingroup$
Once upon a time, this problem let me win a competition :D Here it is another reference: static.elitesecurity.org/uploads/1/9/1987242/…
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 20:01
|
show 4 more comments
0
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4
$begingroup$
cut-the-knot.org/triangle/80-80-20/IndexToClassical.shtml
$endgroup$
– Aretino
Apr 9 '18 at 19:52
$begingroup$
OMG, thanks!!!!
$endgroup$
– greedoid
Apr 9 '18 at 19:53
$begingroup$
Hmm, should I delete this one now?
$endgroup$
– greedoid
Apr 9 '18 at 19:54
$begingroup$
@ChristianF, no, Aretino should turn that comment into an answer :)
$endgroup$
– The Chaz 2.0
Apr 9 '18 at 19:55
1
$begingroup$
Once upon a time, this problem let me win a competition :D Here it is another reference: static.elitesecurity.org/uploads/1/9/1987242/…
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 20:01