Side length of a quadrilateral incribed on a circle
$begingroup$
I've been doing math for 10 years now, yet every so often I get stumped by a "basic" high school question. This is one of those times.
Here's the question:
Part a is easy; we apply the cosine rule to the angle $92^circ$.
I don't understand how to use the sine rule to find $CD$. I managed to find it using the cosine rule; draw a line between $BD$ and use this as the "main" length in the cosine rule. The other two lengths are $5cm$ and $CD$. You end up with a quadratic involving $CD$ that can be solved.
But I don't understand how to use the sine rule to do this. Ideas anyone?
geometry trigonometry circle quadrilateral
asked 2 hours ago
goblingoblin
37.1k1159193
$endgroup$
add a comment |
$begingroup$
I've been doing math for 10 years now, yet every so often I get stumped by a "basic" high school question. This is one of those times.
Here's the question:
Part a is easy; we apply the cosine rule to the angle $92^circ$.
I don't understand how to use the sine rule to find $CD$. I managed to find it using the cosine rule; draw a line between $BD$ and use this as the "main" length in the cosine rule. The other two lengths are $5cm$ and $CD$. You end up with a quadratic involving $CD$ that can be solved.
But I don't understand how to use the sine rule to do this. Ideas anyone?
geometry trigonometry circle quadrilateral
asked 2 hours ago
goblingoblin
37.1k1159193
$endgroup$
add a comment |
$begingroup$
I've been doing math for 10 years now, yet every so often I get stumped by a "basic" high school question. This is one of those times.
Here's the question:
Part a is easy; we apply the cosine rule to the angle $92^circ$.
I don't understand how to use the sine rule to find $CD$. I managed to find it using the cosine rule; draw a line between $BD$ and use this as the "main" length in the cosine rule. The other two lengths are $5cm$ and $CD$. You end up with a quadratic involving $CD$ that can be solved.
But I don't understand how to use the sine rule to do this. Ideas anyone?
geometry trigonometry circle quadrilateral
asked 2 hours ago
goblingoblin
37.1k1159193
$endgroup$
I've been doing math for 10 years now, yet every so often I get stumped by a "basic" high school question. This is one of those times.
Here's the question:
Part a is easy; we apply the cosine rule to the angle $92^circ$.
I don't understand how to use the sine rule to find $CD$. I managed to find it using the cosine rule; draw a line between $BD$ and use this as the "main" length in the cosine rule. The other two lengths are $5cm$ and $CD$. You end up with a quadratic involving $CD$ that can be solved.
But I don't understand how to use the sine rule to do this. Ideas anyone?
geometry trigonometry circle quadrilateral
geometry trigonometry circle quadrilateral
asked 2 hours ago
goblingoblin
37.1k1159193
asked 2 hours ago
goblingoblin
37.1k1159193
asked 2 hours ago
goblingoblin
37.1k1159193
asked 2 hours ago
goblingoblin
37.1k1159193
asked 2 hours ago
goblingoblin
37.1k1159193
37.1k1159193
add a comment |
add a comment |
2 Answers
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$begingroup$
With sines? We need to find $angle CBD$. To do so, we can find $angle BDC$ by the law of sines comparing to $angle DCB$. Then $angle BDC = 180^circ-angle BDC-angle DCB$. It's not particularly convenient, but at least it works.
answered 2 hours ago
jmerryjmerry
15.1k1632
$endgroup$
add a comment |
$begingroup$
You can compute it quite easily.
$measuredangle ABD= measuredangle ACD$ can be computed by sine rule,and $measuredangle CDB = measuredangle CAB$ can be computed in the same way, so you now basically know all the angles of the cyclic quadrilateral.
answered 2 hours ago
Shamim AkhtarShamim Akhtar
145
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Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
With sines? We need to find $angle CBD$. To do so, we can find $angle BDC$ by the law of sines comparing to $angle DCB$. Then $angle BDC = 180^circ-angle BDC-angle DCB$. It's not particularly convenient, but at least it works.
answered 2 hours ago
jmerryjmerry
15.1k1632
$endgroup$
add a comment |
$begingroup$
With sines? We need to find $angle CBD$. To do so, we can find $angle BDC$ by the law of sines comparing to $angle DCB$. Then $angle BDC = 180^circ-angle BDC-angle DCB$. It's not particularly convenient, but at least it works.
answered 2 hours ago
jmerryjmerry
15.1k1632
$endgroup$
add a comment |
$begingroup$
With sines? We need to find $angle CBD$. To do so, we can find $angle BDC$ by the law of sines comparing to $angle DCB$. Then $angle BDC = 180^circ-angle BDC-angle DCB$. It's not particularly convenient, but at least it works.
answered 2 hours ago
jmerryjmerry
15.1k1632
$endgroup$
With sines? We need to find $angle CBD$. To do so, we can find $angle BDC$ by the law of sines comparing to $angle DCB$. Then $angle BDC = 180^circ-angle BDC-angle DCB$. It's not particularly convenient, but at least it works.
answered 2 hours ago
jmerryjmerry
15.1k1632
answered 2 hours ago
jmerryjmerry
15.1k1632
answered 2 hours ago
jmerryjmerry
15.1k1632
answered 2 hours ago
jmerryjmerry
15.1k1632
15.1k1632
add a comment |
add a comment |
$begingroup$
You can compute it quite easily.
$measuredangle ABD= measuredangle ACD$ can be computed by sine rule,and $measuredangle CDB = measuredangle CAB$ can be computed in the same way, so you now basically know all the angles of the cyclic quadrilateral.
answered 2 hours ago
Shamim AkhtarShamim Akhtar
145
New contributor
Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
You can compute it quite easily.
$measuredangle ABD= measuredangle ACD$ can be computed by sine rule,and $measuredangle CDB = measuredangle CAB$ can be computed in the same way, so you now basically know all the angles of the cyclic quadrilateral.
answered 2 hours ago
Shamim AkhtarShamim Akhtar
145
New contributor
Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
You can compute it quite easily.
$measuredangle ABD= measuredangle ACD$ can be computed by sine rule,and $measuredangle CDB = measuredangle CAB$ can be computed in the same way, so you now basically know all the angles of the cyclic quadrilateral.
answered 2 hours ago
Shamim AkhtarShamim Akhtar
145
New contributor
Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You can compute it quite easily.
$measuredangle ABD= measuredangle ACD$ can be computed by sine rule,and $measuredangle CDB = measuredangle CAB$ can be computed in the same way, so you now basically know all the angles of the cyclic quadrilateral.
answered 2 hours ago
Shamim AkhtarShamim Akhtar
145
New contributor
Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Shamim AkhtarShamim Akhtar
145
New contributor
Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
Shamim AkhtarShamim Akhtar
145
answered 2 hours ago
Shamim AkhtarShamim Akhtar
145
145
New contributor
Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Shamim Akhtar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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