Mixing
Two circles with a common tangent
$begingroup$
Find the angle $angle BAC$ in the following picture .
My attempt : I tried to apply the relationships in the both circles between different angles and arcs in many ways but it didn't work . Also joining the point $A$ , $E$ and the line $d$ didn't help (Note that if the intersection point of the $AE$ and the line $d$ is $D$ then $BD = DC$)
geometry circles angle tangent-line
asked Feb 1 at 21:59
S.H.WS.H.W
98511023
$endgroup$
add a comment |
$begingroup$
Find the angle $angle BAC$ in the following picture .
My attempt : I tried to apply the relationships in the both circles between different angles and arcs in many ways but it didn't work . Also joining the point $A$ , $E$ and the line $d$ didn't help (Note that if the intersection point of the $AE$ and the line $d$ is $D$ then $BD = DC$)
geometry circles angle tangent-line
asked Feb 1 at 21:59
S.H.WS.H.W
98511023
$endgroup$
$begingroup$
Why do you think $BD=DC$? Which intersection point $D$ do you have in mind?
$endgroup$
– user376343
Feb 1 at 22:06
$begingroup$
@user376343 See this : math.stackexchange.com/questions/433236/…
$endgroup$
– S.H.W
Feb 1 at 22:07
$begingroup$
The point where $AE$ cuts $d$? Please write it in your question.
$endgroup$
– user376343
Feb 1 at 22:09
1
$begingroup$
@user376343 Thanks , I edited it .
$endgroup$
– S.H.W
Feb 1 at 22:11
add a comment |
$begingroup$
Find the angle $angle BAC$ in the following picture .
My attempt : I tried to apply the relationships in the both circles between different angles and arcs in many ways but it didn't work . Also joining the point $A$ , $E$ and the line $d$ didn't help (Note that if the intersection point of the $AE$ and the line $d$ is $D$ then $BD = DC$)
geometry circles angle tangent-line
asked Feb 1 at 21:59
S.H.WS.H.W
98511023
$endgroup$
Find the angle $angle BAC$ in the following picture .
My attempt : I tried to apply the relationships in the both circles between different angles and arcs in many ways but it didn't work . Also joining the point $A$ , $E$ and the line $d$ didn't help (Note that if the intersection point of the $AE$ and the line $d$ is $D$ then $BD = DC$)
geometry circles angle tangent-line
geometry circles angle tangent-line
asked Feb 1 at 21:59
S.H.WS.H.W
98511023
asked Feb 1 at 21:59
S.H.WS.H.W
98511023
edited Feb 1 at 22:11
S.H.W
asked Feb 1 at 21:59
S.H.WS.H.W
98511023
asked Feb 1 at 21:59
S.H.WS.H.W
98511023
asked Feb 1 at 21:59
S.H.WS.H.W
98511023
98511023
$begingroup$
Why do you think $BD=DC$? Which intersection point $D$ do you have in mind?
$endgroup$
– user376343
Feb 1 at 22:06
$begingroup$
@user376343 See this : math.stackexchange.com/questions/433236/…
$endgroup$
– S.H.W
Feb 1 at 22:07
$begingroup$
The point where $AE$ cuts $d$? Please write it in your question.
$endgroup$
– user376343
Feb 1 at 22:09
1
$begingroup$
@user376343 Thanks , I edited it .
$endgroup$
– S.H.W
Feb 1 at 22:11
add a comment |
$begingroup$
Why do you think $BD=DC$? Which intersection point $D$ do you have in mind?
$endgroup$
– user376343
Feb 1 at 22:06
$begingroup$
@user376343 See this : math.stackexchange.com/questions/433236/…
$endgroup$
– S.H.W
Feb 1 at 22:07
$begingroup$
The point where $AE$ cuts $d$? Please write it in your question.
$endgroup$
– user376343
Feb 1 at 22:09
1
$begingroup$
@user376343 Thanks , I edited it .
$endgroup$
– S.H.W
Feb 1 at 22:11
$begingroup$
Why do you think $BD=DC$? Which intersection point $D$ do you have in mind?
$endgroup$
– user376343
Feb 1 at 22:06
$begingroup$
Why do you think $BD=DC$? Which intersection point $D$ do you have in mind?
$endgroup$
– user376343
Feb 1 at 22:06
$begingroup$
@user376343 See this : math.stackexchange.com/questions/433236/…
$endgroup$
– S.H.W
Feb 1 at 22:07
$begingroup$
@user376343 See this : math.stackexchange.com/questions/433236/…
$endgroup$
– S.H.W
Feb 1 at 22:07
$begingroup$
The point where $AE$ cuts $d$? Please write it in your question.
$endgroup$
– user376343
Feb 1 at 22:09
$begingroup$
The point where $AE$ cuts $d$? Please write it in your question.
$endgroup$
– user376343
Feb 1 at 22:09
1
1
$begingroup$
@user376343 Thanks , I edited it .
$endgroup$
– S.H.W
Feb 1 at 22:11
$begingroup$
@user376343 Thanks , I edited it .
$endgroup$
– S.H.W
Feb 1 at 22:11
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Note that $angle BAE = angle EBC$ because they both correspond to the circular arc $overset{frown}{BE}$
Similarly, $angle CAE = overset{frown}{CE} = angle ECB$
Now, the total internal angle of $triangle ABC$ consists ofbegin{align}
180^{circ} &= color{magenta}{angle BAC} + 30^{circ} + angle EBC + angle ECB + 40^{circ} \
&= color{magenta}{angle BAC} + 30^{circ} + color{magenta}{big(}angle BAE + angle CAEcolor{magenta}{big)} + 40^{circ} \
&= color{magenta}{2angle BAC} + 30^{circ} + 40^{circ}
end{align}
Thus the desired $displaystyleangle BAC = frac{180^{circ} - 30^{circ} - 40^{circ}}2 = 55^{circ}$
answered Feb 2 at 3:32
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– S.H.W
Feb 2 at 3:50
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
Note that $angle BAE = angle EBC$ because they both correspond to the circular arc $overset{frown}{BE}$
Similarly, $angle CAE = overset{frown}{CE} = angle ECB$
Now, the total internal angle of $triangle ABC$ consists ofbegin{align}
180^{circ} &= color{magenta}{angle BAC} + 30^{circ} + angle EBC + angle ECB + 40^{circ} \
&= color{magenta}{angle BAC} + 30^{circ} + color{magenta}{big(}angle BAE + angle CAEcolor{magenta}{big)} + 40^{circ} \
&= color{magenta}{2angle BAC} + 30^{circ} + 40^{circ}
end{align}
Thus the desired $displaystyleangle BAC = frac{180^{circ} - 30^{circ} - 40^{circ}}2 = 55^{circ}$
answered Feb 2 at 3:32
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– S.H.W
Feb 2 at 3:50
add a comment |
$begingroup$
Note that $angle BAE = angle EBC$ because they both correspond to the circular arc $overset{frown}{BE}$
Similarly, $angle CAE = overset{frown}{CE} = angle ECB$
Now, the total internal angle of $triangle ABC$ consists ofbegin{align}
180^{circ} &= color{magenta}{angle BAC} + 30^{circ} + angle EBC + angle ECB + 40^{circ} \
&= color{magenta}{angle BAC} + 30^{circ} + color{magenta}{big(}angle BAE + angle CAEcolor{magenta}{big)} + 40^{circ} \
&= color{magenta}{2angle BAC} + 30^{circ} + 40^{circ}
end{align}
Thus the desired $displaystyleangle BAC = frac{180^{circ} - 30^{circ} - 40^{circ}}2 = 55^{circ}$
answered Feb 2 at 3:32
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– S.H.W
Feb 2 at 3:50
add a comment |
$begingroup$
Note that $angle BAE = angle EBC$ because they both correspond to the circular arc $overset{frown}{BE}$
Similarly, $angle CAE = overset{frown}{CE} = angle ECB$
Now, the total internal angle of $triangle ABC$ consists ofbegin{align}
180^{circ} &= color{magenta}{angle BAC} + 30^{circ} + angle EBC + angle ECB + 40^{circ} \
&= color{magenta}{angle BAC} + 30^{circ} + color{magenta}{big(}angle BAE + angle CAEcolor{magenta}{big)} + 40^{circ} \
&= color{magenta}{2angle BAC} + 30^{circ} + 40^{circ}
end{align}
Thus the desired $displaystyleangle BAC = frac{180^{circ} - 30^{circ} - 40^{circ}}2 = 55^{circ}$
answered Feb 2 at 3:32
Lee David Chung LinLee David Chung Lin
4,47841242
$endgroup$
Note that $angle BAE = angle EBC$ because they both correspond to the circular arc $overset{frown}{BE}$
Similarly, $angle CAE = overset{frown}{CE} = angle ECB$
Now, the total internal angle of $triangle ABC$ consists ofbegin{align}
180^{circ} &= color{magenta}{angle BAC} + 30^{circ} + angle EBC + angle ECB + 40^{circ} \
&= color{magenta}{angle BAC} + 30^{circ} + color{magenta}{big(}angle BAE + angle CAEcolor{magenta}{big)} + 40^{circ} \
&= color{magenta}{2angle BAC} + 30^{circ} + 40^{circ}
end{align}
Thus the desired $displaystyleangle BAC = frac{180^{circ} - 30^{circ} - 40^{circ}}2 = 55^{circ}$
answered Feb 2 at 3:32
Lee David Chung LinLee David Chung Lin
4,47841242
edited Feb 2 at 3:38
answered Feb 2 at 3:32
Lee David Chung LinLee David Chung Lin
4,47841242
answered Feb 2 at 3:32
Lee David Chung LinLee David Chung Lin
4,47841242
answered Feb 2 at 3:32
Lee David Chung LinLee David Chung Lin
4,47841242
4,47841242
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– S.H.W
Feb 2 at 3:50
add a comment |
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– S.H.W
Feb 2 at 3:50
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– S.H.W
Feb 2 at 3:50
$begingroup$
Brilliant! Thanks a lot.
$endgroup$
– S.H.W
Feb 2 at 3:50
add a comment |
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$begingroup$
Why do you think $BD=DC$? Which intersection point $D$ do you have in mind?
$endgroup$
– user376343
Feb 1 at 22:06
$begingroup$
@user376343 See this : math.stackexchange.com/questions/433236/…
$endgroup$
– S.H.W
Feb 1 at 22:07
$begingroup$
The point where $AE$ cuts $d$? Please write it in your question.
$endgroup$
– user376343
Feb 1 at 22:09
1
$begingroup$
@user376343 Thanks , I edited it .
$endgroup$
– S.H.W
Feb 1 at 22:11