CIRCLES (FINDING ANGLES) The 2019 Stack Overflow Developer Survey Results Are InProve $Delta APB $ is equilateral triangleThree circles having centres on the three sides of a triangleGeometry question, prove that $angle APB = frac12 (angle AMB + angle CMD)$Find the angle between two chords passing through points where lines are tangent to the circleQuestion on circles geometryFinding an angle in a figure involving tangent circlesCircle centre point from two angles and circle overalRadius of other circle=?A question about simple circles and trianglesProve two angles on tangents of two circles are equal
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CIRCLES (FINDING ANGLES)
The 2019 Stack Overflow Developer Survey Results Are InProve $Delta APB $ is equilateral triangleThree circles having centres on the three sides of a triangleGeometry question, prove that $angle APB = frac12 (angle AMB + angle CMD)$Find the angle between two chords passing through points where lines are tangent to the circleQuestion on circles geometryFinding an angle in a figure involving tangent circlesCircle centre point from two angles and circle overalRadius of other circle=?A question about simple circles and trianglesProve two angles on tangents of two circles are equal
$begingroup$
In the given figure a semi-circle is drawn with centre M and AB is diameter. If $measuredangle MQN= measuredangle MPN= 10^texto$ and $ measuredangle AMQ=40^texto$, then the measure of $angle PMB$ equals
$$(1)quad20^textoquadquadquadquad(2)quad30^texto$$
My attempts:
geometry euclidean-geometry circles
edited Mar 24 at 8:04
MarianD
2,2611618
asked Mar 24 at 6:19
all about everythingall about everything
112
$endgroup$
add a comment |
$begingroup$
In the given figure a semi-circle is drawn with centre M and AB is diameter. If $measuredangle MQN= measuredangle MPN= 10^texto$ and $ measuredangle AMQ=40^texto$, then the measure of $angle PMB$ equals
$$(1)quad20^textoquadquadquadquad(2)quad30^texto$$
My attempts:
geometry euclidean-geometry circles
edited Mar 24 at 8:04
MarianD
2,2611618
asked Mar 24 at 6:19
all about everythingall about everything
112
$endgroup$
$begingroup$
shall i post it in answers?
$endgroup$
– all about everything
Mar 24 at 6:31
$begingroup$
Post your trying in your starting post.
$endgroup$
– Michael Rozenberg
Mar 24 at 6:36
$begingroup$
Welcome to MSE! Please show us your attempts at the problems, so we can build on it and show you where you went wrong. Thank you, and hope you enjoy being here!
$endgroup$
– BadAtGeometry
Mar 24 at 6:37
1
$begingroup$
i posted my attempts and thank you very much for asking my attempts
$endgroup$
– all about everything
Mar 24 at 6:42
add a comment |
$begingroup$
In the given figure a semi-circle is drawn with centre M and AB is diameter. If $measuredangle MQN= measuredangle MPN= 10^texto$ and $ measuredangle AMQ=40^texto$, then the measure of $angle PMB$ equals
$$(1)quad20^textoquadquadquadquad(2)quad30^texto$$
My attempts:
geometry euclidean-geometry circles
edited Mar 24 at 8:04
MarianD
2,2611618
asked Mar 24 at 6:19
all about everythingall about everything
112
$endgroup$
In the given figure a semi-circle is drawn with centre M and AB is diameter. If $measuredangle MQN= measuredangle MPN= 10^texto$ and $ measuredangle AMQ=40^texto$, then the measure of $angle PMB$ equals
$$(1)quad20^textoquadquadquadquad(2)quad30^texto$$
My attempts:
geometry euclidean-geometry circles
geometry euclidean-geometry circles
edited Mar 24 at 8:04
MarianD
2,2611618
asked Mar 24 at 6:19
all about everythingall about everything
112
edited Mar 24 at 8:04
MarianD
2,2611618
asked Mar 24 at 6:19
all about everythingall about everything
112
edited Mar 24 at 8:04
MarianD
2,2611618
edited Mar 24 at 8:04
MarianD
2,2611618
edited Mar 24 at 8:04
MarianD
2,2611618
2,2611618
asked Mar 24 at 6:19
all about everythingall about everything
112
asked Mar 24 at 6:19
all about everythingall about everything
112
asked Mar 24 at 6:19
all about everythingall about everything
112
112
$begingroup$
shall i post it in answers?
$endgroup$
– all about everything
Mar 24 at 6:31
$begingroup$
Post your trying in your starting post.
$endgroup$
– Michael Rozenberg
Mar 24 at 6:36
$begingroup$
Welcome to MSE! Please show us your attempts at the problems, so we can build on it and show you where you went wrong. Thank you, and hope you enjoy being here!
$endgroup$
– BadAtGeometry
Mar 24 at 6:37
1
$begingroup$
i posted my attempts and thank you very much for asking my attempts
$endgroup$
– all about everything
Mar 24 at 6:42
add a comment |
$begingroup$
shall i post it in answers?
$endgroup$
– all about everything
Mar 24 at 6:31
$begingroup$
Post your trying in your starting post.
$endgroup$
– Michael Rozenberg
Mar 24 at 6:36
$begingroup$
Welcome to MSE! Please show us your attempts at the problems, so we can build on it and show you where you went wrong. Thank you, and hope you enjoy being here!
$endgroup$
– BadAtGeometry
Mar 24 at 6:37
1
$begingroup$
i posted my attempts and thank you very much for asking my attempts
$endgroup$
– all about everything
Mar 24 at 6:42
$begingroup$
shall i post it in answers?
$endgroup$
– all about everything
Mar 24 at 6:31
$begingroup$
shall i post it in answers?
$endgroup$
– all about everything
Mar 24 at 6:31
$begingroup$
Post your trying in your starting post.
$endgroup$
– Michael Rozenberg
Mar 24 at 6:36
$begingroup$
Post your trying in your starting post.
$endgroup$
– Michael Rozenberg
Mar 24 at 6:36
$begingroup$
Welcome to MSE! Please show us your attempts at the problems, so we can build on it and show you where you went wrong. Thank you, and hope you enjoy being here!
$endgroup$
– BadAtGeometry
Mar 24 at 6:37
$begingroup$
Welcome to MSE! Please show us your attempts at the problems, so we can build on it and show you where you went wrong. Thank you, and hope you enjoy being here!
$endgroup$
– BadAtGeometry
Mar 24 at 6:37
1
1
$begingroup$
i posted my attempts and thank you very much for asking my attempts
$endgroup$
– all about everything
Mar 24 at 6:42
$begingroup$
i posted my attempts and thank you very much for asking my attempts
$endgroup$
– all about everything
Mar 24 at 6:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $measuredangle PMN=x$.
Since $measuredangle MQN=measuredangle MPN$, we obtain that $MQPN$ is cyclic.
Thus, $$measuredangle NQP=measuredangle NMP=x,$$ which says $$measuredangle QPN=10^circ+x+10^circ=20^circ+x.$$
Id est, $$20^circ+x=40^circ,$$ which gives $$x=20^circ.$$
answered Mar 24 at 6:35
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
I think PMB is 20°. Here is my explanation:
The quadrilateral MNPQ has to be cyclic because angle MQN=angle MPN. So angle QMN +angle NPQ =180.
Therefore, angle QPN= 40°. So, angle QPM= 40-10=30°.
Now,angle QMA=40°. SO angle QPA=20°. (Half of QMA).So angle APM=10°. now angle APB=90° So angle NPB=70°. now in traingle MPB, angle MPB=MBP. so angle PMB=20°
answered Mar 24 at 6:45
TojrahTojrah
4036
$endgroup$
$begingroup$
thanks , nice explaination!
$endgroup$
– all about everything
Mar 24 at 6:53
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $measuredangle PMN=x$.
Since $measuredangle MQN=measuredangle MPN$, we obtain that $MQPN$ is cyclic.
Thus, $$measuredangle NQP=measuredangle NMP=x,$$ which says $$measuredangle QPN=10^circ+x+10^circ=20^circ+x.$$
Id est, $$20^circ+x=40^circ,$$ which gives $$x=20^circ.$$
answered Mar 24 at 6:35
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Let $measuredangle PMN=x$.
Since $measuredangle MQN=measuredangle MPN$, we obtain that $MQPN$ is cyclic.
Thus, $$measuredangle NQP=measuredangle NMP=x,$$ which says $$measuredangle QPN=10^circ+x+10^circ=20^circ+x.$$
Id est, $$20^circ+x=40^circ,$$ which gives $$x=20^circ.$$
answered Mar 24 at 6:35
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
Let $measuredangle PMN=x$.
Since $measuredangle MQN=measuredangle MPN$, we obtain that $MQPN$ is cyclic.
Thus, $$measuredangle NQP=measuredangle NMP=x,$$ which says $$measuredangle QPN=10^circ+x+10^circ=20^circ+x.$$
Id est, $$20^circ+x=40^circ,$$ which gives $$x=20^circ.$$
answered Mar 24 at 6:35
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
Let $measuredangle PMN=x$.
Since $measuredangle MQN=measuredangle MPN$, we obtain that $MQPN$ is cyclic.
Thus, $$measuredangle NQP=measuredangle NMP=x,$$ which says $$measuredangle QPN=10^circ+x+10^circ=20^circ+x.$$
Id est, $$20^circ+x=40^circ,$$ which gives $$x=20^circ.$$
answered Mar 24 at 6:35
Michael RozenbergMichael Rozenberg
110k1896201
answered Mar 24 at 6:35
Michael RozenbergMichael Rozenberg
110k1896201
answered Mar 24 at 6:35
Michael RozenbergMichael Rozenberg
110k1896201
answered Mar 24 at 6:35
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
add a comment |
add a comment |
$begingroup$
I think PMB is 20°. Here is my explanation:
The quadrilateral MNPQ has to be cyclic because angle MQN=angle MPN. So angle QMN +angle NPQ =180.
Therefore, angle QPN= 40°. So, angle QPM= 40-10=30°.
Now,angle QMA=40°. SO angle QPA=20°. (Half of QMA).So angle APM=10°. now angle APB=90° So angle NPB=70°. now in traingle MPB, angle MPB=MBP. so angle PMB=20°
answered Mar 24 at 6:45
TojrahTojrah
4036
$endgroup$
$begingroup$
thanks , nice explaination!
$endgroup$
– all about everything
Mar 24 at 6:53
add a comment |
$begingroup$
I think PMB is 20°. Here is my explanation:
The quadrilateral MNPQ has to be cyclic because angle MQN=angle MPN. So angle QMN +angle NPQ =180.
Therefore, angle QPN= 40°. So, angle QPM= 40-10=30°.
Now,angle QMA=40°. SO angle QPA=20°. (Half of QMA).So angle APM=10°. now angle APB=90° So angle NPB=70°. now in traingle MPB, angle MPB=MBP. so angle PMB=20°
answered Mar 24 at 6:45
TojrahTojrah
4036
$endgroup$
$begingroup$
thanks , nice explaination!
$endgroup$
– all about everything
Mar 24 at 6:53
add a comment |
$begingroup$
I think PMB is 20°. Here is my explanation:
The quadrilateral MNPQ has to be cyclic because angle MQN=angle MPN. So angle QMN +angle NPQ =180.
Therefore, angle QPN= 40°. So, angle QPM= 40-10=30°.
Now,angle QMA=40°. SO angle QPA=20°. (Half of QMA).So angle APM=10°. now angle APB=90° So angle NPB=70°. now in traingle MPB, angle MPB=MBP. so angle PMB=20°
answered Mar 24 at 6:45
TojrahTojrah
4036
$endgroup$
I think PMB is 20°. Here is my explanation:
The quadrilateral MNPQ has to be cyclic because angle MQN=angle MPN. So angle QMN +angle NPQ =180.
Therefore, angle QPN= 40°. So, angle QPM= 40-10=30°.
Now,angle QMA=40°. SO angle QPA=20°. (Half of QMA).So angle APM=10°. now angle APB=90° So angle NPB=70°. now in traingle MPB, angle MPB=MBP. so angle PMB=20°
answered Mar 24 at 6:45
TojrahTojrah
4036
answered Mar 24 at 6:45
TojrahTojrah
4036
answered Mar 24 at 6:45
TojrahTojrah
4036
answered Mar 24 at 6:45
TojrahTojrah
4036
4036
$begingroup$
thanks , nice explaination!
$endgroup$
– all about everything
Mar 24 at 6:53
add a comment |
$begingroup$
thanks , nice explaination!
$endgroup$
– all about everything
Mar 24 at 6:53
$begingroup$
thanks , nice explaination!
$endgroup$
– all about everything
Mar 24 at 6:53
$begingroup$
thanks , nice explaination!
$endgroup$
– all about everything
Mar 24 at 6:53
add a comment |
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$begingroup$
shall i post it in answers?
$endgroup$
– all about everything
Mar 24 at 6:31
$begingroup$
Post your trying in your starting post.
$endgroup$
– Michael Rozenberg
Mar 24 at 6:36
$begingroup$
Welcome to MSE! Please show us your attempts at the problems, so we can build on it and show you where you went wrong. Thank you, and hope you enjoy being here!
$endgroup$
– BadAtGeometry
Mar 24 at 6:37
1
$begingroup$
i posted my attempts and thank you very much for asking my attempts
$endgroup$
– all about everything
Mar 24 at 6:42