Find the angle marked [closed]
$begingroup$
OABC is a square. Find the angle $x$. I have tried to draw a right triangle inside the square.
geometry
edited Jan 26 at 11:01
Maria Mazur
46.7k1260120
asked Jan 25 at 21:43
EnzoEnzo
19917
$endgroup$
closed as off-topic by Eevee Trainer, Did, Jyrki Lahtonen, Shailesh, darij grinberg Feb 8 at 2:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, Jyrki Lahtonen, Shailesh, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
OABC is a square. Find the angle $x$. I have tried to draw a right triangle inside the square.
geometry
edited Jan 26 at 11:01
Maria Mazur
46.7k1260120
asked Jan 25 at 21:43
EnzoEnzo
19917
$endgroup$
closed as off-topic by Eevee Trainer, Did, Jyrki Lahtonen, Shailesh, darij grinberg Feb 8 at 2:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, Jyrki Lahtonen, Shailesh, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Triangle DOB is isosceles and the angle DOB cuts 1/8 of the circle.
$endgroup$
– B. Goddard
Jan 25 at 21:46
add a comment |
$begingroup$
OABC is a square. Find the angle $x$. I have tried to draw a right triangle inside the square.
geometry
edited Jan 26 at 11:01
Maria Mazur
46.7k1260120
asked Jan 25 at 21:43
EnzoEnzo
19917
$endgroup$
OABC is a square. Find the angle $x$. I have tried to draw a right triangle inside the square.
geometry
geometry
edited Jan 26 at 11:01
Maria Mazur
46.7k1260120
asked Jan 25 at 21:43
EnzoEnzo
19917
edited Jan 26 at 11:01
Maria Mazur
46.7k1260120
asked Jan 25 at 21:43
EnzoEnzo
19917
edited Jan 26 at 11:01
Maria Mazur
46.7k1260120
edited Jan 26 at 11:01
Maria Mazur
46.7k1260120
edited Jan 26 at 11:01
Maria Mazur
46.7k1260120
46.7k1260120
asked Jan 25 at 21:43
EnzoEnzo
19917
asked Jan 25 at 21:43
EnzoEnzo
19917
asked Jan 25 at 21:43
EnzoEnzo
19917
19917
closed as off-topic by Eevee Trainer, Did, Jyrki Lahtonen, Shailesh, darij grinberg Feb 8 at 2:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, Jyrki Lahtonen, Shailesh, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, Did, Jyrki Lahtonen, Shailesh, darij grinberg Feb 8 at 2:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, Jyrki Lahtonen, Shailesh, darij grinberg
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Triangle DOB is isosceles and the angle DOB cuts 1/8 of the circle.
$endgroup$
– B. Goddard
Jan 25 at 21:46
add a comment |
3
$begingroup$
Triangle DOB is isosceles and the angle DOB cuts 1/8 of the circle.
$endgroup$
– B. Goddard
Jan 25 at 21:46
3
3
$begingroup$
Triangle DOB is isosceles and the angle DOB cuts 1/8 of the circle.
$endgroup$
– B. Goddard
Jan 25 at 21:46
$begingroup$
Triangle DOB is isosceles and the angle DOB cuts 1/8 of the circle.
$endgroup$
– B. Goddard
Jan 25 at 21:46
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint
The radius of the quarter circle (in terms of the side length of the square) is $$sqrt{2} s$$
Where $s$ is the length of a side of square $ABCO$
What does that make the length of $DC$?
Edit
We have $DC = sqrt{2}s - s = s(sqrt{2} - 1)$. By using trigonometry, we can say that $$tan{x} = frac{s}{s(sqrt{2} - 1)} = frac{1}{sqrt{2} - 1}$$
Think you can take it from here? A calculator will probably be needed to solve for $x$. A perhaps nicer approach is given in the comments by B. Goddard and a calculator is not required
answered Jan 25 at 21:47
WaveXWaveX
2,7622722
$endgroup$
$begingroup$
Please clarify what does "s" stand for, thanks.
$endgroup$
– NoChance
Jan 25 at 21:49
$begingroup$
It is the side length of square $ABCO$. However it doesn't matter the exact value as it will be cancelled out in the trig...
$endgroup$
– WaveX
Jan 25 at 21:51
$begingroup$
@WaveX Is it 90 degrees? I did not get your hint.
$endgroup$
– Enzo
Jan 25 at 21:52
$begingroup$
We do not know the side lenght here.
$endgroup$
– Enzo
Jan 25 at 21:58
$begingroup$
$OD$ is also a radius of the quarter circle, so it must be $sqrt{2} s$. And we defined $OC$ to be of length $s$. By subtraction, we can get $DC = $OD-OC$. Then we can use trigonometry to figure out the angle. The side lengths will cancel out so don't worry about being left with an $s$ in the final answer
$endgroup$
– WaveX
Jan 25 at 22:00
|
show 2 more comments
$begingroup$
If you draw a line between O and B you get the picture below.
The line segment OB and the line segment OD, each is a radius, so OB=OD
This means that the triangle ODB has 2 equal sides
This means that the angle OBD=angle ODB, call that "x"....(1).
Now the line OB bisects the 90-degree angle COA. So the angle BOC (or BOD) is 45 degrees.
We know that BOD+OBD+ODB=180
$45+2*x=180$
So, x=OBD=ODB=$(180-45)/2=67.5$ degrees.
answered Jan 25 at 22:16
NoChanceNoChance
3,76721221
$endgroup$
add a comment |
$begingroup$
Draw a tangent at $D$ and let $X$ be on it (right from $D$).
Then by tangent-chord theorem we have $$angle BDX = {1over 2}angle BOD = 22,5^{circ}$$
So $$angle ODB= 67,5^{circ}$$
answered Jan 25 at 22:15
Maria MazurMaria Mazur
46.7k1260120
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
The radius of the quarter circle (in terms of the side length of the square) is $$sqrt{2} s$$
Where $s$ is the length of a side of square $ABCO$
What does that make the length of $DC$?
Edit
We have $DC = sqrt{2}s - s = s(sqrt{2} - 1)$. By using trigonometry, we can say that $$tan{x} = frac{s}{s(sqrt{2} - 1)} = frac{1}{sqrt{2} - 1}$$
Think you can take it from here? A calculator will probably be needed to solve for $x$. A perhaps nicer approach is given in the comments by B. Goddard and a calculator is not required
answered Jan 25 at 21:47
WaveXWaveX
2,7622722
$endgroup$
$begingroup$
Please clarify what does "s" stand for, thanks.
$endgroup$
– NoChance
Jan 25 at 21:49
$begingroup$
It is the side length of square $ABCO$. However it doesn't matter the exact value as it will be cancelled out in the trig...
$endgroup$
– WaveX
Jan 25 at 21:51
$begingroup$
@WaveX Is it 90 degrees? I did not get your hint.
$endgroup$
– Enzo
Jan 25 at 21:52
$begingroup$
We do not know the side lenght here.
$endgroup$
– Enzo
Jan 25 at 21:58
$begingroup$
$OD$ is also a radius of the quarter circle, so it must be $sqrt{2} s$. And we defined $OC$ to be of length $s$. By subtraction, we can get $DC = $OD-OC$. Then we can use trigonometry to figure out the angle. The side lengths will cancel out so don't worry about being left with an $s$ in the final answer
$endgroup$
– WaveX
Jan 25 at 22:00
|
show 2 more comments
$begingroup$
Hint
The radius of the quarter circle (in terms of the side length of the square) is $$sqrt{2} s$$
Where $s$ is the length of a side of square $ABCO$
What does that make the length of $DC$?
Edit
We have $DC = sqrt{2}s - s = s(sqrt{2} - 1)$. By using trigonometry, we can say that $$tan{x} = frac{s}{s(sqrt{2} - 1)} = frac{1}{sqrt{2} - 1}$$
Think you can take it from here? A calculator will probably be needed to solve for $x$. A perhaps nicer approach is given in the comments by B. Goddard and a calculator is not required
answered Jan 25 at 21:47
WaveXWaveX
2,7622722
$endgroup$
$begingroup$
Please clarify what does "s" stand for, thanks.
$endgroup$
– NoChance
Jan 25 at 21:49
$begingroup$
It is the side length of square $ABCO$. However it doesn't matter the exact value as it will be cancelled out in the trig...
$endgroup$
– WaveX
Jan 25 at 21:51
$begingroup$
@WaveX Is it 90 degrees? I did not get your hint.
$endgroup$
– Enzo
Jan 25 at 21:52
$begingroup$
We do not know the side lenght here.
$endgroup$
– Enzo
Jan 25 at 21:58
$begingroup$
$OD$ is also a radius of the quarter circle, so it must be $sqrt{2} s$. And we defined $OC$ to be of length $s$. By subtraction, we can get $DC = $OD-OC$. Then we can use trigonometry to figure out the angle. The side lengths will cancel out so don't worry about being left with an $s$ in the final answer
$endgroup$
– WaveX
Jan 25 at 22:00
|
show 2 more comments
$begingroup$
Hint
The radius of the quarter circle (in terms of the side length of the square) is $$sqrt{2} s$$
Where $s$ is the length of a side of square $ABCO$
What does that make the length of $DC$?
Edit
We have $DC = sqrt{2}s - s = s(sqrt{2} - 1)$. By using trigonometry, we can say that $$tan{x} = frac{s}{s(sqrt{2} - 1)} = frac{1}{sqrt{2} - 1}$$
Think you can take it from here? A calculator will probably be needed to solve for $x$. A perhaps nicer approach is given in the comments by B. Goddard and a calculator is not required
answered Jan 25 at 21:47
WaveXWaveX
2,7622722
$endgroup$
Hint
The radius of the quarter circle (in terms of the side length of the square) is $$sqrt{2} s$$
Where $s$ is the length of a side of square $ABCO$
What does that make the length of $DC$?
Edit
We have $DC = sqrt{2}s - s = s(sqrt{2} - 1)$. By using trigonometry, we can say that $$tan{x} = frac{s}{s(sqrt{2} - 1)} = frac{1}{sqrt{2} - 1}$$
Think you can take it from here? A calculator will probably be needed to solve for $x$. A perhaps nicer approach is given in the comments by B. Goddard and a calculator is not required
answered Jan 25 at 21:47
WaveXWaveX
2,7622722
edited Jan 25 at 22:10
answered Jan 25 at 21:47
WaveXWaveX
2,7622722
answered Jan 25 at 21:47
WaveXWaveX
2,7622722
answered Jan 25 at 21:47
WaveXWaveX
2,7622722
2,7622722
$begingroup$
Please clarify what does "s" stand for, thanks.
$endgroup$
– NoChance
Jan 25 at 21:49
$begingroup$
It is the side length of square $ABCO$. However it doesn't matter the exact value as it will be cancelled out in the trig...
$endgroup$
– WaveX
Jan 25 at 21:51
$begingroup$
@WaveX Is it 90 degrees? I did not get your hint.
$endgroup$
– Enzo
Jan 25 at 21:52
$begingroup$
We do not know the side lenght here.
$endgroup$
– Enzo
Jan 25 at 21:58
$begingroup$
$OD$ is also a radius of the quarter circle, so it must be $sqrt{2} s$. And we defined $OC$ to be of length $s$. By subtraction, we can get $DC = $OD-OC$. Then we can use trigonometry to figure out the angle. The side lengths will cancel out so don't worry about being left with an $s$ in the final answer
$endgroup$
– WaveX
Jan 25 at 22:00
|
show 2 more comments
$begingroup$
Please clarify what does "s" stand for, thanks.
$endgroup$
– NoChance
Jan 25 at 21:49
$begingroup$
It is the side length of square $ABCO$. However it doesn't matter the exact value as it will be cancelled out in the trig...
$endgroup$
– WaveX
Jan 25 at 21:51
$begingroup$
@WaveX Is it 90 degrees? I did not get your hint.
$endgroup$
– Enzo
Jan 25 at 21:52
$begingroup$
We do not know the side lenght here.
$endgroup$
– Enzo
Jan 25 at 21:58
$begingroup$
$OD$ is also a radius of the quarter circle, so it must be $sqrt{2} s$. And we defined $OC$ to be of length $s$. By subtraction, we can get $DC = $OD-OC$. Then we can use trigonometry to figure out the angle. The side lengths will cancel out so don't worry about being left with an $s$ in the final answer
$endgroup$
– WaveX
Jan 25 at 22:00
$begingroup$
Please clarify what does "s" stand for, thanks.
$endgroup$
– NoChance
Jan 25 at 21:49
$begingroup$
Please clarify what does "s" stand for, thanks.
$endgroup$
– NoChance
Jan 25 at 21:49
$begingroup$
It is the side length of square $ABCO$. However it doesn't matter the exact value as it will be cancelled out in the trig...
$endgroup$
– WaveX
Jan 25 at 21:51
$begingroup$
It is the side length of square $ABCO$. However it doesn't matter the exact value as it will be cancelled out in the trig...
$endgroup$
– WaveX
Jan 25 at 21:51
$begingroup$
@WaveX Is it 90 degrees? I did not get your hint.
$endgroup$
– Enzo
Jan 25 at 21:52
$begingroup$
@WaveX Is it 90 degrees? I did not get your hint.
$endgroup$
– Enzo
Jan 25 at 21:52
$begingroup$
We do not know the side lenght here.
$endgroup$
– Enzo
Jan 25 at 21:58
$begingroup$
We do not know the side lenght here.
$endgroup$
– Enzo
Jan 25 at 21:58
$begingroup$
$OD$ is also a radius of the quarter circle, so it must be $sqrt{2} s$. And we defined $OC$ to be of length $s$. By subtraction, we can get $DC = $OD-OC$. Then we can use trigonometry to figure out the angle. The side lengths will cancel out so don't worry about being left with an $s$ in the final answer
$endgroup$
– WaveX
Jan 25 at 22:00
$begingroup$
$OD$ is also a radius of the quarter circle, so it must be $sqrt{2} s$. And we defined $OC$ to be of length $s$. By subtraction, we can get $DC = $OD-OC$. Then we can use trigonometry to figure out the angle. The side lengths will cancel out so don't worry about being left with an $s$ in the final answer
$endgroup$
– WaveX
Jan 25 at 22:00
|
show 2 more comments
$begingroup$
If you draw a line between O and B you get the picture below.
The line segment OB and the line segment OD, each is a radius, so OB=OD
This means that the triangle ODB has 2 equal sides
This means that the angle OBD=angle ODB, call that "x"....(1).
Now the line OB bisects the 90-degree angle COA. So the angle BOC (or BOD) is 45 degrees.
We know that BOD+OBD+ODB=180
$45+2*x=180$
So, x=OBD=ODB=$(180-45)/2=67.5$ degrees.
answered Jan 25 at 22:16
NoChanceNoChance
3,76721221
$endgroup$
add a comment |
$begingroup$
If you draw a line between O and B you get the picture below.
The line segment OB and the line segment OD, each is a radius, so OB=OD
This means that the triangle ODB has 2 equal sides
This means that the angle OBD=angle ODB, call that "x"....(1).
Now the line OB bisects the 90-degree angle COA. So the angle BOC (or BOD) is 45 degrees.
We know that BOD+OBD+ODB=180
$45+2*x=180$
So, x=OBD=ODB=$(180-45)/2=67.5$ degrees.
answered Jan 25 at 22:16
NoChanceNoChance
3,76721221
$endgroup$
add a comment |
$begingroup$
If you draw a line between O and B you get the picture below.
The line segment OB and the line segment OD, each is a radius, so OB=OD
This means that the triangle ODB has 2 equal sides
This means that the angle OBD=angle ODB, call that "x"....(1).
Now the line OB bisects the 90-degree angle COA. So the angle BOC (or BOD) is 45 degrees.
We know that BOD+OBD+ODB=180
$45+2*x=180$
So, x=OBD=ODB=$(180-45)/2=67.5$ degrees.
answered Jan 25 at 22:16
NoChanceNoChance
3,76721221
$endgroup$
If you draw a line between O and B you get the picture below.
The line segment OB and the line segment OD, each is a radius, so OB=OD
This means that the triangle ODB has 2 equal sides
This means that the angle OBD=angle ODB, call that "x"....(1).
Now the line OB bisects the 90-degree angle COA. So the angle BOC (or BOD) is 45 degrees.
We know that BOD+OBD+ODB=180
$45+2*x=180$
So, x=OBD=ODB=$(180-45)/2=67.5$ degrees.
answered Jan 25 at 22:16
NoChanceNoChance
3,76721221
answered Jan 25 at 22:16
NoChanceNoChance
3,76721221
answered Jan 25 at 22:16
NoChanceNoChance
3,76721221
answered Jan 25 at 22:16
NoChanceNoChance
3,76721221
3,76721221
add a comment |
add a comment |
$begingroup$
Draw a tangent at $D$ and let $X$ be on it (right from $D$).
Then by tangent-chord theorem we have $$angle BDX = {1over 2}angle BOD = 22,5^{circ}$$
So $$angle ODB= 67,5^{circ}$$
answered Jan 25 at 22:15
Maria MazurMaria Mazur
46.7k1260120
$endgroup$
add a comment |
$begingroup$
Draw a tangent at $D$ and let $X$ be on it (right from $D$).
Then by tangent-chord theorem we have $$angle BDX = {1over 2}angle BOD = 22,5^{circ}$$
So $$angle ODB= 67,5^{circ}$$
answered Jan 25 at 22:15
Maria MazurMaria Mazur
46.7k1260120
$endgroup$
add a comment |
$begingroup$
Draw a tangent at $D$ and let $X$ be on it (right from $D$).
Then by tangent-chord theorem we have $$angle BDX = {1over 2}angle BOD = 22,5^{circ}$$
So $$angle ODB= 67,5^{circ}$$
answered Jan 25 at 22:15
Maria MazurMaria Mazur
46.7k1260120
$endgroup$
Draw a tangent at $D$ and let $X$ be on it (right from $D$).
Then by tangent-chord theorem we have $$angle BDX = {1over 2}angle BOD = 22,5^{circ}$$
So $$angle ODB= 67,5^{circ}$$
answered Jan 25 at 22:15
Maria MazurMaria Mazur
46.7k1260120
answered Jan 25 at 22:15
Maria MazurMaria Mazur
46.7k1260120
answered Jan 25 at 22:15
Maria MazurMaria Mazur
46.7k1260120
answered Jan 25 at 22:15
Maria MazurMaria Mazur
46.7k1260120
46.7k1260120
add a comment |
add a comment |
3
$begingroup$
Triangle DOB is isosceles and the angle DOB cuts 1/8 of the circle.
$endgroup$
– B. Goddard
Jan 25 at 21:46