Mixing
How to find an angle of a triangle that has three points of tangency? [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
1
My question is about angles in the circumference.
In the figure shown, the circumferences are exterior tangents. Being $C$, $A$ and $Q$ tangency points. If the measure of the $ABC$ arc is $290$. Find the excess of $120$ over $x$.
The truth does not find a way to solve it.
geometry circle tangent-line
edited Jul 30 at 21:58
Xander Henderson
13.1k83150
asked Jul 30 at 21:57
Payo
61
closed as off-topic by Xander Henderson, Batominovski, amWhy, Isaac Browne, Leucippus Jul 31 at 3:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Batominovski, amWhy, Isaac Browne, Leucippus
add a comment |Â
up vote
0
down vote
favorite
1
My question is about angles in the circumference.
In the figure shown, the circumferences are exterior tangents. Being $C$, $A$ and $Q$ tangency points. If the measure of the $ABC$ arc is $290$. Find the excess of $120$ over $x$.
The truth does not find a way to solve it.
geometry circle tangent-line
edited Jul 30 at 21:58
Xander Henderson
13.1k83150
asked Jul 30 at 21:57
Payo
61
closed as off-topic by Xander Henderson, Batominovski, amWhy, Isaac Browne, Leucippus Jul 31 at 3:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Batominovski, amWhy, Isaac Browne, Leucippus
1
It would be helpful (and it is likely that your question would be better received) if you could provide some additional context. What have you tried to solve this problem? What tools (theorems, definitions, etc) do you have access to? Why are you interested in this problem? Why should anyone else be interested in this problem?
– Xander Henderson
Jul 30 at 22:01
How is B defined?
– Doug M
Jul 30 at 22:02
1
@DougM The point $B$ is probably just there so as to not confuse to which arc is being referred (the minor arc $AC$ or the major arc $AC$). So, with $B$, we know that $ABC$ is the major arc. What does "excess of $120$ over $x$" mean, though? Simply $120-x$?
– Batominovski
Jul 30 at 22:06
add a comment |Â
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
My question is about angles in the circumference.
In the figure shown, the circumferences are exterior tangents. Being $C$, $A$ and $Q$ tangency points. If the measure of the $ABC$ arc is $290$. Find the excess of $120$ over $x$.
The truth does not find a way to solve it.
geometry circle tangent-line
edited Jul 30 at 21:58
Xander Henderson
13.1k83150
asked Jul 30 at 21:57
Payo
61
My question is about angles in the circumference.
In the figure shown, the circumferences are exterior tangents. Being $C$, $A$ and $Q$ tangency points. If the measure of the $ABC$ arc is $290$. Find the excess of $120$ over $x$.
The truth does not find a way to solve it.
geometry circle tangent-line
edited Jul 30 at 21:58
Xander Henderson
13.1k83150
asked Jul 30 at 21:57
Payo
61
edited Jul 30 at 21:58
Xander Henderson
13.1k83150
edited Jul 30 at 21:58
Xander Henderson
13.1k83150
edited Jul 30 at 21:58
Xander Henderson
13.1k83150
13.1k83150
asked Jul 30 at 21:57
Payo
61
asked Jul 30 at 21:57
Payo
61
asked Jul 30 at 21:57
Payo
61
61
closed as off-topic by Xander Henderson, Batominovski, amWhy, Isaac Browne, Leucippus Jul 31 at 3:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Batominovski, amWhy, Isaac Browne, Leucippus
closed as off-topic by Xander Henderson, Batominovski, amWhy, Isaac Browne, Leucippus Jul 31 at 3:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Xander Henderson, Batominovski, amWhy, Isaac Browne, Leucippus
1
It would be helpful (and it is likely that your question would be better received) if you could provide some additional context. What have you tried to solve this problem? What tools (theorems, definitions, etc) do you have access to? Why are you interested in this problem? Why should anyone else be interested in this problem?
– Xander Henderson
Jul 30 at 22:01
How is B defined?
– Doug M
Jul 30 at 22:02
1
@DougM The point $B$ is probably just there so as to not confuse to which arc is being referred (the minor arc $AC$ or the major arc $AC$). So, with $B$, we know that $ABC$ is the major arc. What does "excess of $120$ over $x$" mean, though? Simply $120-x$?
– Batominovski
Jul 30 at 22:06
add a comment |Â
1
It would be helpful (and it is likely that your question would be better received) if you could provide some additional context. What have you tried to solve this problem? What tools (theorems, definitions, etc) do you have access to? Why are you interested in this problem? Why should anyone else be interested in this problem?
– Xander Henderson
Jul 30 at 22:01
How is B defined?
– Doug M
Jul 30 at 22:02
1
@DougM The point $B$ is probably just there so as to not confuse to which arc is being referred (the minor arc $AC$ or the major arc $AC$). So, with $B$, we know that $ABC$ is the major arc. What does "excess of $120$ over $x$" mean, though? Simply $120-x$?
– Batominovski
Jul 30 at 22:06
1
1
It would be helpful (and it is likely that your question would be better received) if you could provide some additional context. What have you tried to solve this problem? What tools (theorems, definitions, etc) do you have access to? Why are you interested in this problem? Why should anyone else be interested in this problem?
– Xander Henderson
Jul 30 at 22:01
It would be helpful (and it is likely that your question would be better received) if you could provide some additional context. What have you tried to solve this problem? What tools (theorems, definitions, etc) do you have access to? Why are you interested in this problem? Why should anyone else be interested in this problem?
– Xander Henderson
Jul 30 at 22:01
How is B defined?
– Doug M
Jul 30 at 22:02
How is B defined?
– Doug M
Jul 30 at 22:02
1
1
@DougM The point $B$ is probably just there so as to not confuse to which arc is being referred (the minor arc $AC$ or the major arc $AC$). So, with $B$, we know that $ABC$ is the major arc. What does "excess of $120$ over $x$" mean, though? Simply $120-x$?
– Batominovski
Jul 30 at 22:06
@DougM The point $B$ is probably just there so as to not confuse to which arc is being referred (the minor arc $AC$ or the major arc $AC$). So, with $B$, we know that $ABC$ is the major arc. What does "excess of $120$ over $x$" mean, though? Simply $120-x$?
– Batominovski
Jul 30 at 22:06
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Hint: We have $angle ABC=frac360^circ-290^circ2$ and $angle ABC=angle CAQ$. Plus, $angle CQA=angle APQ$. You can also show that $angle ACQ=90^circ$.
answered Jul 30 at 22:10
Batominovski
22.8k22776
add a comment |Â
up vote
2
down vote
I assume that $P$ is the antipodal point of $Q$ and $C$ is a point where the two circles are tangent to each other.
By the Tangent-Chord Theorem we have that $angle CAQ = angle CBA = 35^circ$
Finally as $triangle PQA$ is right triangle we get that $x = 55^circ$, so the excess is $65^circ$
answered Jul 30 at 22:11
Stefan4024
27.8k52974
Is it well known that $angle PQA=90^circ$? There is a short proof, but it is nontrivial enough. I find it easier to show that $angle ACQ=90^circ$ instead.
– Batominovski
Jul 30 at 22:18
@Batominovski Yeah, eventually it reduces to proving that $angle ACQ = 90$. Here's my proof: Let $A'$ be the intersection of $QC$ with the bigger circle. Then by power of point we have $AC cdot AP = AQ^2 = QC cdot QA'$. This gives that the triangles $AA'C$ and $PQC$ are similar and hence $angle A'AC = x$. Also $angle AA'C = angle CAQ$, so we get that $AA'C$ is similar with $PAQ$ and subsequentially with $ACQ$. So we have $angle ACQ = angle ACA'$ and as their sum is 180 we get what we want.
– Stefan4024
Jul 30 at 23:12
I have a homothetic proof without involving $angle ACQ$. The homothety about $C$ sending the circle $ABC$ to the circle $PQC$ with $A$ mapped to $P$ also send the tangent line $AQ$ of the circle $ABC$ to the tangent line of the circle $PQC$ at $P$. The two lines must be parallel, and since $AQ$ touches the circle $PQC$ at $Q$, we conclude that $PQ$ is a diameter of the circle $PQC$, whence it is perpendicular to $AQ$.
– Batominovski
Jul 30 at 23:15
add a comment |Â
up vote
1
down vote
$〇＋×ï¼Â90°$ and $ACQsim AQP$ make this problem easy. 
answered Jul 31 at 0:04
Takahiro Waki
1,964520
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: We have $angle ABC=frac360^circ-290^circ2$ and $angle ABC=angle CAQ$. Plus, $angle CQA=angle APQ$. You can also show that $angle ACQ=90^circ$.
answered Jul 30 at 22:10
Batominovski
22.8k22776
add a comment |Â
up vote
2
down vote
accepted
Hint: We have $angle ABC=frac360^circ-290^circ2$ and $angle ABC=angle CAQ$. Plus, $angle CQA=angle APQ$. You can also show that $angle ACQ=90^circ$.
answered Jul 30 at 22:10
Batominovski
22.8k22776
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: We have $angle ABC=frac360^circ-290^circ2$ and $angle ABC=angle CAQ$. Plus, $angle CQA=angle APQ$. You can also show that $angle ACQ=90^circ$.
answered Jul 30 at 22:10
Batominovski
22.8k22776
Hint: We have $angle ABC=frac360^circ-290^circ2$ and $angle ABC=angle CAQ$. Plus, $angle CQA=angle APQ$. You can also show that $angle ACQ=90^circ$.
answered Jul 30 at 22:10
Batominovski
22.8k22776
edited Jul 30 at 22:16
answered Jul 30 at 22:10
Batominovski
22.8k22776
answered Jul 30 at 22:10
Batominovski
22.8k22776
answered Jul 30 at 22:10
Batominovski
22.8k22776
22.8k22776
add a comment |Â
add a comment |Â
up vote
2
down vote
I assume that $P$ is the antipodal point of $Q$ and $C$ is a point where the two circles are tangent to each other.
By the Tangent-Chord Theorem we have that $angle CAQ = angle CBA = 35^circ$
Finally as $triangle PQA$ is right triangle we get that $x = 55^circ$, so the excess is $65^circ$
answered Jul 30 at 22:11
Stefan4024
27.8k52974
Is it well known that $angle PQA=90^circ$? There is a short proof, but it is nontrivial enough. I find it easier to show that $angle ACQ=90^circ$ instead.
– Batominovski
Jul 30 at 22:18
@Batominovski Yeah, eventually it reduces to proving that $angle ACQ = 90$. Here's my proof: Let $A'$ be the intersection of $QC$ with the bigger circle. Then by power of point we have $AC cdot AP = AQ^2 = QC cdot QA'$. This gives that the triangles $AA'C$ and $PQC$ are similar and hence $angle A'AC = x$. Also $angle AA'C = angle CAQ$, so we get that $AA'C$ is similar with $PAQ$ and subsequentially with $ACQ$. So we have $angle ACQ = angle ACA'$ and as their sum is 180 we get what we want.
– Stefan4024
Jul 30 at 23:12
I have a homothetic proof without involving $angle ACQ$. The homothety about $C$ sending the circle $ABC$ to the circle $PQC$ with $A$ mapped to $P$ also send the tangent line $AQ$ of the circle $ABC$ to the tangent line of the circle $PQC$ at $P$. The two lines must be parallel, and since $AQ$ touches the circle $PQC$ at $Q$, we conclude that $PQ$ is a diameter of the circle $PQC$, whence it is perpendicular to $AQ$.
– Batominovski
Jul 30 at 23:15
add a comment |Â
up vote
2
down vote
I assume that $P$ is the antipodal point of $Q$ and $C$ is a point where the two circles are tangent to each other.
By the Tangent-Chord Theorem we have that $angle CAQ = angle CBA = 35^circ$
Finally as $triangle PQA$ is right triangle we get that $x = 55^circ$, so the excess is $65^circ$
answered Jul 30 at 22:11
Stefan4024
27.8k52974
Is it well known that $angle PQA=90^circ$? There is a short proof, but it is nontrivial enough. I find it easier to show that $angle ACQ=90^circ$ instead.
– Batominovski
Jul 30 at 22:18
@Batominovski Yeah, eventually it reduces to proving that $angle ACQ = 90$. Here's my proof: Let $A'$ be the intersection of $QC$ with the bigger circle. Then by power of point we have $AC cdot AP = AQ^2 = QC cdot QA'$. This gives that the triangles $AA'C$ and $PQC$ are similar and hence $angle A'AC = x$. Also $angle AA'C = angle CAQ$, so we get that $AA'C$ is similar with $PAQ$ and subsequentially with $ACQ$. So we have $angle ACQ = angle ACA'$ and as their sum is 180 we get what we want.
– Stefan4024
Jul 30 at 23:12
I have a homothetic proof without involving $angle ACQ$. The homothety about $C$ sending the circle $ABC$ to the circle $PQC$ with $A$ mapped to $P$ also send the tangent line $AQ$ of the circle $ABC$ to the tangent line of the circle $PQC$ at $P$. The two lines must be parallel, and since $AQ$ touches the circle $PQC$ at $Q$, we conclude that $PQ$ is a diameter of the circle $PQC$, whence it is perpendicular to $AQ$.
– Batominovski
Jul 30 at 23:15
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I assume that $P$ is the antipodal point of $Q$ and $C$ is a point where the two circles are tangent to each other.
By the Tangent-Chord Theorem we have that $angle CAQ = angle CBA = 35^circ$
Finally as $triangle PQA$ is right triangle we get that $x = 55^circ$, so the excess is $65^circ$
answered Jul 30 at 22:11
Stefan4024
27.8k52974
I assume that $P$ is the antipodal point of $Q$ and $C$ is a point where the two circles are tangent to each other.
By the Tangent-Chord Theorem we have that $angle CAQ = angle CBA = 35^circ$
Finally as $triangle PQA$ is right triangle we get that $x = 55^circ$, so the excess is $65^circ$
answered Jul 30 at 22:11
Stefan4024
27.8k52974
answered Jul 30 at 22:11
Stefan4024
27.8k52974
answered Jul 30 at 22:11
Stefan4024
27.8k52974
answered Jul 30 at 22:11
Stefan4024
27.8k52974
27.8k52974
Is it well known that $angle PQA=90^circ$? There is a short proof, but it is nontrivial enough. I find it easier to show that $angle ACQ=90^circ$ instead.
– Batominovski
Jul 30 at 22:18
@Batominovski Yeah, eventually it reduces to proving that $angle ACQ = 90$. Here's my proof: Let $A'$ be the intersection of $QC$ with the bigger circle. Then by power of point we have $AC cdot AP = AQ^2 = QC cdot QA'$. This gives that the triangles $AA'C$ and $PQC$ are similar and hence $angle A'AC = x$. Also $angle AA'C = angle CAQ$, so we get that $AA'C$ is similar with $PAQ$ and subsequentially with $ACQ$. So we have $angle ACQ = angle ACA'$ and as their sum is 180 we get what we want.
– Stefan4024
Jul 30 at 23:12
I have a homothetic proof without involving $angle ACQ$. The homothety about $C$ sending the circle $ABC$ to the circle $PQC$ with $A$ mapped to $P$ also send the tangent line $AQ$ of the circle $ABC$ to the tangent line of the circle $PQC$ at $P$. The two lines must be parallel, and since $AQ$ touches the circle $PQC$ at $Q$, we conclude that $PQ$ is a diameter of the circle $PQC$, whence it is perpendicular to $AQ$.
– Batominovski
Jul 30 at 23:15
add a comment |Â
Is it well known that $angle PQA=90^circ$? There is a short proof, but it is nontrivial enough. I find it easier to show that $angle ACQ=90^circ$ instead.
– Batominovski
Jul 30 at 22:18
@Batominovski Yeah, eventually it reduces to proving that $angle ACQ = 90$. Here's my proof: Let $A'$ be the intersection of $QC$ with the bigger circle. Then by power of point we have $AC cdot AP = AQ^2 = QC cdot QA'$. This gives that the triangles $AA'C$ and $PQC$ are similar and hence $angle A'AC = x$. Also $angle AA'C = angle CAQ$, so we get that $AA'C$ is similar with $PAQ$ and subsequentially with $ACQ$. So we have $angle ACQ = angle ACA'$ and as their sum is 180 we get what we want.
– Stefan4024
Jul 30 at 23:12
I have a homothetic proof without involving $angle ACQ$. The homothety about $C$ sending the circle $ABC$ to the circle $PQC$ with $A$ mapped to $P$ also send the tangent line $AQ$ of the circle $ABC$ to the tangent line of the circle $PQC$ at $P$. The two lines must be parallel, and since $AQ$ touches the circle $PQC$ at $Q$, we conclude that $PQ$ is a diameter of the circle $PQC$, whence it is perpendicular to $AQ$.
– Batominovski
Jul 30 at 23:15
Is it well known that $angle PQA=90^circ$? There is a short proof, but it is nontrivial enough. I find it easier to show that $angle ACQ=90^circ$ instead.
– Batominovski
Jul 30 at 22:18
Is it well known that $angle PQA=90^circ$? There is a short proof, but it is nontrivial enough. I find it easier to show that $angle ACQ=90^circ$ instead.
– Batominovski
Jul 30 at 22:18
@Batominovski Yeah, eventually it reduces to proving that $angle ACQ = 90$. Here's my proof: Let $A'$ be the intersection of $QC$ with the bigger circle. Then by power of point we have $AC cdot AP = AQ^2 = QC cdot QA'$. This gives that the triangles $AA'C$ and $PQC$ are similar and hence $angle A'AC = x$. Also $angle AA'C = angle CAQ$, so we get that $AA'C$ is similar with $PAQ$ and subsequentially with $ACQ$. So we have $angle ACQ = angle ACA'$ and as their sum is 180 we get what we want.
– Stefan4024
Jul 30 at 23:12
@Batominovski Yeah, eventually it reduces to proving that $angle ACQ = 90$. Here's my proof: Let $A'$ be the intersection of $QC$ with the bigger circle. Then by power of point we have $AC cdot AP = AQ^2 = QC cdot QA'$. This gives that the triangles $AA'C$ and $PQC$ are similar and hence $angle A'AC = x$. Also $angle AA'C = angle CAQ$, so we get that $AA'C$ is similar with $PAQ$ and subsequentially with $ACQ$. So we have $angle ACQ = angle ACA'$ and as their sum is 180 we get what we want.
– Stefan4024
Jul 30 at 23:12
I have a homothetic proof without involving $angle ACQ$. The homothety about $C$ sending the circle $ABC$ to the circle $PQC$ with $A$ mapped to $P$ also send the tangent line $AQ$ of the circle $ABC$ to the tangent line of the circle $PQC$ at $P$. The two lines must be parallel, and since $AQ$ touches the circle $PQC$ at $Q$, we conclude that $PQ$ is a diameter of the circle $PQC$, whence it is perpendicular to $AQ$.
– Batominovski
Jul 30 at 23:15
I have a homothetic proof without involving $angle ACQ$. The homothety about $C$ sending the circle $ABC$ to the circle $PQC$ with $A$ mapped to $P$ also send the tangent line $AQ$ of the circle $ABC$ to the tangent line of the circle $PQC$ at $P$. The two lines must be parallel, and since $AQ$ touches the circle $PQC$ at $Q$, we conclude that $PQ$ is a diameter of the circle $PQC$, whence it is perpendicular to $AQ$.
– Batominovski
Jul 30 at 23:15
add a comment |Â
up vote
1
down vote
$〇＋×ï¼Â90°$ and $ACQsim AQP$ make this problem easy.
answered Jul 31 at 0:04
Takahiro Waki
1,964520
add a comment |Â
up vote
1
down vote
$〇＋×ï¼Â90°$ and $ACQsim AQP$ make this problem easy.
answered Jul 31 at 0:04
Takahiro Waki
1,964520
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$〇＋×ï¼Â90°$ and $ACQsim AQP$ make this problem easy.
answered Jul 31 at 0:04
Takahiro Waki
1,964520
$〇＋×ï¼Â90°$ and $ACQsim AQP$ make this problem easy.
answered Jul 31 at 0:04
Takahiro Waki
1,964520
answered Jul 31 at 0:04
Takahiro Waki
1,964520
answered Jul 31 at 0:04
Takahiro Waki
1,964520
answered Jul 31 at 0:04
Takahiro Waki
1,964520
1,964520
add a comment |Â
add a comment |Â
1
It would be helpful (and it is likely that your question would be better received) if you could provide some additional context. What have you tried to solve this problem? What tools (theorems, definitions, etc) do you have access to? Why are you interested in this problem? Why should anyone else be interested in this problem?
– Xander Henderson
Jul 30 at 22:01
How is B defined?
– Doug M
Jul 30 at 22:02
1
@DougM The point $B$ is probably just there so as to not confuse to which arc is being referred (the minor arc $AC$ or the major arc $AC$). So, with $B$, we know that $ABC$ is the major arc. What does "excess of $120$ over $x$" mean, though? Simply $120-x$?
– Batominovski
Jul 30 at 22:06