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How to find the sum of the sides of a polygon whose one vertex goes from the north of a circle and the other comes from the east in its perimeter?
Clash Royale CLAN TAG#URR8PPP
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The problem is as follows:
In figure 1. there is a circle as shown. The radius is equal to 10 inches and its center is labeled with the letter O. If $measuredangle PC=30^circ$. $textrmFind AB+BC$.
The existing alternatives in my book are:
- $3left( sqrt2+sqrt6right)$
- $4left( sqrt6-sqrt2right)$
- $5left( sqrt3-sqrt2right)$
- $5left( sqrt3+sqrt2right)$
- $5left( sqrt2+sqrt6right)$
After analyzing the drawing the figure from below shows all all the relationships which I could found and it is summarized as follows:
The triangle $textrmCOP$ is isosceles since it shares the same side from the radius of the circle and since $measuredangle PC=30^circ$, then all is left to do is to apply the identity which it says that the sum of inner angles in a triangle must equate to $180^circ$.
$$2x+30^circ=180^circ$$
$$x=frac150^circ2=75^circ$$
Since $measuredangle OCP = measuredangle OPC$, its supplementary angle would become:
$$180^circ-75^circ=105^circ$$
Since it is given from the problem:
$$measuredangle COA = 90^circ$$
therefore its complementary angle with $measuredangle COP = 30^circ$ would become into:
$$measuredangle POA = 60^circ$$
Since $PO = OA$ this would also make another isosceles triangle and by recurring to the previous identity:
$$2x+60^circ=180^circ$$
$$x=frac180^circ-60^circ2=60^circ$$
Therefore the triangle POA is an equilateral one so,
$$textrmPA=10 inches$$
As $measuredangle OPB = 105 ^circ$ and $measuredangle OPA = 60^circ$ then its difference is:
$measuredangle APB = 45^circ$.
From this its easy to note that $measuredangle PAB = 45^circ$.
Since the vertex $textrmB$ of the triangle $textrmABP$ is $measuredangle = 90 ^circ$. I did identified a special right triangle with the form $45^circ-45^circ-90^circ$ or $textrmk, k,,ksqrt2$.
By equating the newly found side $textrmPA = 10 inches$ to $ksqrt2$ this is transformed into:
$$ksqrt2 = 10$$
$$k = frac10sqrt2$$
From this is established that:
$$AB = frac10sqrt2$$
Since we have $textrmAB$ we also know $textrmPB$ as $AB = PB = frac10sqrt2$
Therefore all that is left to do is to find $textrmCP$ as $CP+PB = BC$
To find $CP$ I used cosines law as follows:
$$a^2=b^2+c^2-2bc,cos A$$
Being a, b and c the sides of a triangle ABC and A the opposing angle from the side taken as a reference in the left side of the equation.
In this case
$$(CP)^2= 10^2+10^2-2(10)(10)cos30^circ$$
$$(CP)^2= 10^2 left(1+1-2left(fracsqrt32right)right)$$
$$CP = 10 sqrt 2-sqrt3$$
Therefore $CP = 10 sqrt 2-sqrt3$ and we have all the parts so the rest is just adding them up.
$$CP+PB= BC = 10 sqrt 2-sqrt3 + frac10sqrt2$$
$$AB= frac10sqrt2$$
$$AB + BC = frac10sqrt2 + 10 sqrt 2-sqrt3 + frac10sqrt2$$
And that's how far I went, but from then on I don't know if what I did was correct or did I missed something? as my answer doesn't appear within the alternatives.
The best I could come up with by simplifying was:
$$frac10sqrt22+10sqrt2-sqrt3+frac10sqrt22$$
$$10sqrt2+10sqrt2-sqrt3$$
$$10left(sqrt2+sqrt2-sqrt3right)$$
and, that's it. But it doesn't seem to be in the choices given. Can somebody help me to find if did I do something wrong?. If a drawing is necessary please include one as I'm not savvy enough to notice these things easily.
algebra-precalculus geometry euclidean-geometry
asked 11 hours ago
Chris Steinbeck Bell
682314
add a comment |Â
up vote
8
down vote
favorite
The problem is as follows:
In figure 1. there is a circle as shown. The radius is equal to 10 inches and its center is labeled with the letter O. If $measuredangle PC=30^circ$. $textrmFind AB+BC$.
The existing alternatives in my book are:
- $3left( sqrt2+sqrt6right)$
- $4left( sqrt6-sqrt2right)$
- $5left( sqrt3-sqrt2right)$
- $5left( sqrt3+sqrt2right)$
- $5left( sqrt2+sqrt6right)$
After analyzing the drawing the figure from below shows all all the relationships which I could found and it is summarized as follows:
The triangle $textrmCOP$ is isosceles since it shares the same side from the radius of the circle and since $measuredangle PC=30^circ$, then all is left to do is to apply the identity which it says that the sum of inner angles in a triangle must equate to $180^circ$.
$$2x+30^circ=180^circ$$
$$x=frac150^circ2=75^circ$$
Since $measuredangle OCP = measuredangle OPC$, its supplementary angle would become:
$$180^circ-75^circ=105^circ$$
Since it is given from the problem:
$$measuredangle COA = 90^circ$$
therefore its complementary angle with $measuredangle COP = 30^circ$ would become into:
$$measuredangle POA = 60^circ$$
Since $PO = OA$ this would also make another isosceles triangle and by recurring to the previous identity:
$$2x+60^circ=180^circ$$
$$x=frac180^circ-60^circ2=60^circ$$
Therefore the triangle POA is an equilateral one so,
$$textrmPA=10 inches$$
As $measuredangle OPB = 105 ^circ$ and $measuredangle OPA = 60^circ$ then its difference is:
$measuredangle APB = 45^circ$.
From this its easy to note that $measuredangle PAB = 45^circ$.
Since the vertex $textrmB$ of the triangle $textrmABP$ is $measuredangle = 90 ^circ$. I did identified a special right triangle with the form $45^circ-45^circ-90^circ$ or $textrmk, k,,ksqrt2$.
By equating the newly found side $textrmPA = 10 inches$ to $ksqrt2$ this is transformed into:
$$ksqrt2 = 10$$
$$k = frac10sqrt2$$
From this is established that:
$$AB = frac10sqrt2$$
Since we have $textrmAB$ we also know $textrmPB$ as $AB = PB = frac10sqrt2$
Therefore all that is left to do is to find $textrmCP$ as $CP+PB = BC$
To find $CP$ I used cosines law as follows:
$$a^2=b^2+c^2-2bc,cos A$$
Being a, b and c the sides of a triangle ABC and A the opposing angle from the side taken as a reference in the left side of the equation.
In this case
$$(CP)^2= 10^2+10^2-2(10)(10)cos30^circ$$
$$(CP)^2= 10^2 left(1+1-2left(fracsqrt32right)right)$$
$$CP = 10 sqrt 2-sqrt3$$
Therefore $CP = 10 sqrt 2-sqrt3$ and we have all the parts so the rest is just adding them up.
$$CP+PB= BC = 10 sqrt 2-sqrt3 + frac10sqrt2$$
$$AB= frac10sqrt2$$
$$AB + BC = frac10sqrt2 + 10 sqrt 2-sqrt3 + frac10sqrt2$$
And that's how far I went, but from then on I don't know if what I did was correct or did I missed something? as my answer doesn't appear within the alternatives.
The best I could come up with by simplifying was:
$$frac10sqrt22+10sqrt2-sqrt3+frac10sqrt22$$
$$10sqrt2+10sqrt2-sqrt3$$
$$10left(sqrt2+sqrt2-sqrt3right)$$
and, that's it. But it doesn't seem to be in the choices given. Can somebody help me to find if did I do something wrong?. If a drawing is necessary please include one as I'm not savvy enough to notice these things easily.
algebra-precalculus geometry euclidean-geometry
asked 11 hours ago
Chris Steinbeck Bell
682314
1
If $∡PC=30º$ then I think $angle AOP=60º$ since the angle to the center is twice as the angle to the side.
– abc...
11 hours ago
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
The problem is as follows:
In figure 1. there is a circle as shown. The radius is equal to 10 inches and its center is labeled with the letter O. If $measuredangle PC=30^circ$. $textrmFind AB+BC$.
The existing alternatives in my book are:
- $3left( sqrt2+sqrt6right)$
- $4left( sqrt6-sqrt2right)$
- $5left( sqrt3-sqrt2right)$
- $5left( sqrt3+sqrt2right)$
- $5left( sqrt2+sqrt6right)$
After analyzing the drawing the figure from below shows all all the relationships which I could found and it is summarized as follows:
The triangle $textrmCOP$ is isosceles since it shares the same side from the radius of the circle and since $measuredangle PC=30^circ$, then all is left to do is to apply the identity which it says that the sum of inner angles in a triangle must equate to $180^circ$.
$$2x+30^circ=180^circ$$
$$x=frac150^circ2=75^circ$$
Since $measuredangle OCP = measuredangle OPC$, its supplementary angle would become:
$$180^circ-75^circ=105^circ$$
Since it is given from the problem:
$$measuredangle COA = 90^circ$$
therefore its complementary angle with $measuredangle COP = 30^circ$ would become into:
$$measuredangle POA = 60^circ$$
Since $PO = OA$ this would also make another isosceles triangle and by recurring to the previous identity:
$$2x+60^circ=180^circ$$
$$x=frac180^circ-60^circ2=60^circ$$
Therefore the triangle POA is an equilateral one so,
$$textrmPA=10 inches$$
As $measuredangle OPB = 105 ^circ$ and $measuredangle OPA = 60^circ$ then its difference is:
$measuredangle APB = 45^circ$.
From this its easy to note that $measuredangle PAB = 45^circ$.
Since the vertex $textrmB$ of the triangle $textrmABP$ is $measuredangle = 90 ^circ$. I did identified a special right triangle with the form $45^circ-45^circ-90^circ$ or $textrmk, k,,ksqrt2$.
By equating the newly found side $textrmPA = 10 inches$ to $ksqrt2$ this is transformed into:
$$ksqrt2 = 10$$
$$k = frac10sqrt2$$
From this is established that:
$$AB = frac10sqrt2$$
Since we have $textrmAB$ we also know $textrmPB$ as $AB = PB = frac10sqrt2$
Therefore all that is left to do is to find $textrmCP$ as $CP+PB = BC$
To find $CP$ I used cosines law as follows:
$$a^2=b^2+c^2-2bc,cos A$$
Being a, b and c the sides of a triangle ABC and A the opposing angle from the side taken as a reference in the left side of the equation.
In this case
$$(CP)^2= 10^2+10^2-2(10)(10)cos30^circ$$
$$(CP)^2= 10^2 left(1+1-2left(fracsqrt32right)right)$$
$$CP = 10 sqrt 2-sqrt3$$
Therefore $CP = 10 sqrt 2-sqrt3$ and we have all the parts so the rest is just adding them up.
$$CP+PB= BC = 10 sqrt 2-sqrt3 + frac10sqrt2$$
$$AB= frac10sqrt2$$
$$AB + BC = frac10sqrt2 + 10 sqrt 2-sqrt3 + frac10sqrt2$$
And that's how far I went, but from then on I don't know if what I did was correct or did I missed something? as my answer doesn't appear within the alternatives.
The best I could come up with by simplifying was:
$$frac10sqrt22+10sqrt2-sqrt3+frac10sqrt22$$
$$10sqrt2+10sqrt2-sqrt3$$
$$10left(sqrt2+sqrt2-sqrt3right)$$
and, that's it. But it doesn't seem to be in the choices given. Can somebody help me to find if did I do something wrong?. If a drawing is necessary please include one as I'm not savvy enough to notice these things easily.
algebra-precalculus geometry euclidean-geometry
asked 11 hours ago
Chris Steinbeck Bell
682314
The problem is as follows:
In figure 1. there is a circle as shown. The radius is equal to 10 inches and its center is labeled with the letter O. If $measuredangle PC=30^circ$. $textrmFind AB+BC$.
The existing alternatives in my book are:
- $3left( sqrt2+sqrt6right)$
- $4left( sqrt6-sqrt2right)$
- $5left( sqrt3-sqrt2right)$
- $5left( sqrt3+sqrt2right)$
- $5left( sqrt2+sqrt6right)$
After analyzing the drawing the figure from below shows all all the relationships which I could found and it is summarized as follows:
The triangle $textrmCOP$ is isosceles since it shares the same side from the radius of the circle and since $measuredangle PC=30^circ$, then all is left to do is to apply the identity which it says that the sum of inner angles in a triangle must equate to $180^circ$.
$$2x+30^circ=180^circ$$
$$x=frac150^circ2=75^circ$$
Since $measuredangle OCP = measuredangle OPC$, its supplementary angle would become:
$$180^circ-75^circ=105^circ$$
Since it is given from the problem:
$$measuredangle COA = 90^circ$$
therefore its complementary angle with $measuredangle COP = 30^circ$ would become into:
$$measuredangle POA = 60^circ$$
Since $PO = OA$ this would also make another isosceles triangle and by recurring to the previous identity:
$$2x+60^circ=180^circ$$
$$x=frac180^circ-60^circ2=60^circ$$
Therefore the triangle POA is an equilateral one so,
$$textrmPA=10 inches$$
As $measuredangle OPB = 105 ^circ$ and $measuredangle OPA = 60^circ$ then its difference is:
$measuredangle APB = 45^circ$.
From this its easy to note that $measuredangle PAB = 45^circ$.
Since the vertex $textrmB$ of the triangle $textrmABP$ is $measuredangle = 90 ^circ$. I did identified a special right triangle with the form $45^circ-45^circ-90^circ$ or $textrmk, k,,ksqrt2$.
By equating the newly found side $textrmPA = 10 inches$ to $ksqrt2$ this is transformed into:
$$ksqrt2 = 10$$
$$k = frac10sqrt2$$
From this is established that:
$$AB = frac10sqrt2$$
Since we have $textrmAB$ we also know $textrmPB$ as $AB = PB = frac10sqrt2$
Therefore all that is left to do is to find $textrmCP$ as $CP+PB = BC$
To find $CP$ I used cosines law as follows:
$$a^2=b^2+c^2-2bc,cos A$$
Being a, b and c the sides of a triangle ABC and A the opposing angle from the side taken as a reference in the left side of the equation.
In this case
$$(CP)^2= 10^2+10^2-2(10)(10)cos30^circ$$
$$(CP)^2= 10^2 left(1+1-2left(fracsqrt32right)right)$$
$$CP = 10 sqrt 2-sqrt3$$
Therefore $CP = 10 sqrt 2-sqrt3$ and we have all the parts so the rest is just adding them up.
$$CP+PB= BC = 10 sqrt 2-sqrt3 + frac10sqrt2$$
$$AB= frac10sqrt2$$
$$AB + BC = frac10sqrt2 + 10 sqrt 2-sqrt3 + frac10sqrt2$$
And that's how far I went, but from then on I don't know if what I did was correct or did I missed something? as my answer doesn't appear within the alternatives.
The best I could come up with by simplifying was:
$$frac10sqrt22+10sqrt2-sqrt3+frac10sqrt22$$
$$10sqrt2+10sqrt2-sqrt3$$
$$10left(sqrt2+sqrt2-sqrt3right)$$
and, that's it. But it doesn't seem to be in the choices given. Can somebody help me to find if did I do something wrong?. If a drawing is necessary please include one as I'm not savvy enough to notice these things easily.
algebra-precalculus geometry euclidean-geometry
algebra-precalculus geometry euclidean-geometry
asked 11 hours ago
Chris Steinbeck Bell
682314
asked 11 hours ago
Chris Steinbeck Bell
682314
edited 11 hours ago
asked 11 hours ago
Chris Steinbeck Bell
682314
asked 11 hours ago
Chris Steinbeck Bell
682314
asked 11 hours ago
Chris Steinbeck Bell
682314
682314
1
If $∡PC=30º$ then I think $angle AOP=60º$ since the angle to the center is twice as the angle to the side.
– abc...
11 hours ago
add a comment |Â
1
If $∡PC=30º$ then I think $angle AOP=60º$ since the angle to the center is twice as the angle to the side.
– abc...
11 hours ago
1
1
If $∡PC=30º$ then I think $angle AOP=60º$ since the angle to the center is twice as the angle to the side.
– abc...
11 hours ago
If $∡PC=30º$ then I think $angle AOP=60º$ since the angle to the center is twice as the angle to the side.
– abc...
11 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
Assuming your work is correct so far, I think we may be able to simplify further.
$$sqrt2-sqrt3=sqrtfrac4-2sqrt32=fracsqrt3-1sqrt2$$.
Now let's see if that helps.
$$10left(sqrt2+sqrt2-sqrt3right)=5sqrt2left(2+sqrt3-1right)=5(sqrt2+sqrt6)$$
Again, assuming everything you've done is correct, the last answer is the solution.
answered 11 hours ago
Mike
11.7k31642
add a comment |Â
up vote
3
down vote
Notice that $$angle ACB=angle OCB-angle OCA=75^o-45^o=30^o,$$and $$AC=10sqrt2.$$Thus $$AB+BC=ACcdot(sin angle ACB+cos angle ACB)=5(sqrt2+sqrt6).$$
answered 8 hours ago
mengdie1982
3,683216
add a comment |Â
up vote
0
down vote
I have checked your work and it is correct.
Your answer is correct and $$10left(sqrt2+sqrt2-sqrt3 right)=5left( sqrt2+sqrt6right)=19.31851653$$
answered 11 hours ago
Mohammad Riazi-Kermani
31k41853
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Assuming your work is correct so far, I think we may be able to simplify further.
$$sqrt2-sqrt3=sqrtfrac4-2sqrt32=fracsqrt3-1sqrt2$$.
Now let's see if that helps.
$$10left(sqrt2+sqrt2-sqrt3right)=5sqrt2left(2+sqrt3-1right)=5(sqrt2+sqrt6)$$
Again, assuming everything you've done is correct, the last answer is the solution.
answered 11 hours ago
Mike
11.7k31642
add a comment |Â
up vote
3
down vote
Assuming your work is correct so far, I think we may be able to simplify further.
$$sqrt2-sqrt3=sqrtfrac4-2sqrt32=fracsqrt3-1sqrt2$$.
Now let's see if that helps.
$$10left(sqrt2+sqrt2-sqrt3right)=5sqrt2left(2+sqrt3-1right)=5(sqrt2+sqrt6)$$
Again, assuming everything you've done is correct, the last answer is the solution.
answered 11 hours ago
Mike
11.7k31642
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Assuming your work is correct so far, I think we may be able to simplify further.
$$sqrt2-sqrt3=sqrtfrac4-2sqrt32=fracsqrt3-1sqrt2$$.
Now let's see if that helps.
$$10left(sqrt2+sqrt2-sqrt3right)=5sqrt2left(2+sqrt3-1right)=5(sqrt2+sqrt6)$$
Again, assuming everything you've done is correct, the last answer is the solution.
answered 11 hours ago
Mike
11.7k31642
Assuming your work is correct so far, I think we may be able to simplify further.
$$sqrt2-sqrt3=sqrtfrac4-2sqrt32=fracsqrt3-1sqrt2$$.
Now let's see if that helps.
$$10left(sqrt2+sqrt2-sqrt3right)=5sqrt2left(2+sqrt3-1right)=5(sqrt2+sqrt6)$$
Again, assuming everything you've done is correct, the last answer is the solution.
answered 11 hours ago
Mike
11.7k31642
answered 11 hours ago
Mike
11.7k31642
answered 11 hours ago
Mike
11.7k31642
answered 11 hours ago
Mike
11.7k31642
11.7k31642
add a comment |Â
add a comment |Â
up vote
3
down vote
Notice that $$angle ACB=angle OCB-angle OCA=75^o-45^o=30^o,$$and $$AC=10sqrt2.$$Thus $$AB+BC=ACcdot(sin angle ACB+cos angle ACB)=5(sqrt2+sqrt6).$$
answered 8 hours ago
mengdie1982
3,683216
add a comment |Â
up vote
3
down vote
Notice that $$angle ACB=angle OCB-angle OCA=75^o-45^o=30^o,$$and $$AC=10sqrt2.$$Thus $$AB+BC=ACcdot(sin angle ACB+cos angle ACB)=5(sqrt2+sqrt6).$$
answered 8 hours ago
mengdie1982
3,683216
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Notice that $$angle ACB=angle OCB-angle OCA=75^o-45^o=30^o,$$and $$AC=10sqrt2.$$Thus $$AB+BC=ACcdot(sin angle ACB+cos angle ACB)=5(sqrt2+sqrt6).$$
answered 8 hours ago
mengdie1982
3,683216
Notice that $$angle ACB=angle OCB-angle OCA=75^o-45^o=30^o,$$and $$AC=10sqrt2.$$Thus $$AB+BC=ACcdot(sin angle ACB+cos angle ACB)=5(sqrt2+sqrt6).$$
answered 8 hours ago
mengdie1982
3,683216
answered 8 hours ago
mengdie1982
3,683216
answered 8 hours ago
mengdie1982
3,683216
answered 8 hours ago
mengdie1982
3,683216
3,683216
add a comment |Â
add a comment |Â
up vote
0
down vote
I have checked your work and it is correct.
Your answer is correct and $$10left(sqrt2+sqrt2-sqrt3 right)=5left( sqrt2+sqrt6right)=19.31851653$$
answered 11 hours ago
Mohammad Riazi-Kermani
31k41853
add a comment |Â
up vote
0
down vote
I have checked your work and it is correct.
Your answer is correct and $$10left(sqrt2+sqrt2-sqrt3 right)=5left( sqrt2+sqrt6right)=19.31851653$$
answered 11 hours ago
Mohammad Riazi-Kermani
31k41853
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I have checked your work and it is correct.
Your answer is correct and $$10left(sqrt2+sqrt2-sqrt3 right)=5left( sqrt2+sqrt6right)=19.31851653$$
answered 11 hours ago
Mohammad Riazi-Kermani
31k41853
I have checked your work and it is correct.
Your answer is correct and $$10left(sqrt2+sqrt2-sqrt3 right)=5left( sqrt2+sqrt6right)=19.31851653$$
answered 11 hours ago
Mohammad Riazi-Kermani
31k41853
answered 11 hours ago
Mohammad Riazi-Kermani
31k41853
answered 11 hours ago
Mohammad Riazi-Kermani
31k41853
answered 11 hours ago
Mohammad Riazi-Kermani
31k41853
31k41853
add a comment |Â
add a comment |Â
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1
If $∡PC=30º$ then I think $angle AOP=60º$ since the angle to the center is twice as the angle to the side.
– abc...
11 hours ago