How to compute $BC$ in this triangle?
$begingroup$
Given the lengths of $AF$, $AE$, $AD$, $AC$, How can I calculate $BC$?
I tried to calculate $FD = sqrt{AF^2 - AD^2}$,
$DE = sqrt{AE^2 - AD^2}$,
$FE = DE + FD$,
$FC = AC - AF$.
But I don't know what I can do with all these lengths to compute $BC$. Any hints are appreciated.
geometry euclidean-geometry
asked Dec 29 '18 at 9:03
Mohamed MagdyMohamed Magdy
31127
$endgroup$
add a comment |
$begingroup$
Given the lengths of $AF$, $AE$, $AD$, $AC$, How can I calculate $BC$?
I tried to calculate $FD = sqrt{AF^2 - AD^2}$,
$DE = sqrt{AE^2 - AD^2}$,
$FE = DE + FD$,
$FC = AC - AF$.
But I don't know what I can do with all these lengths to compute $BC$. Any hints are appreciated.
geometry euclidean-geometry
asked Dec 29 '18 at 9:03
Mohamed MagdyMohamed Magdy
31127
$endgroup$
add a comment |
$begingroup$
Given the lengths of $AF$, $AE$, $AD$, $AC$, How can I calculate $BC$?
I tried to calculate $FD = sqrt{AF^2 - AD^2}$,
$DE = sqrt{AE^2 - AD^2}$,
$FE = DE + FD$,
$FC = AC - AF$.
But I don't know what I can do with all these lengths to compute $BC$. Any hints are appreciated.
geometry euclidean-geometry
asked Dec 29 '18 at 9:03
Mohamed MagdyMohamed Magdy
31127
$endgroup$
Given the lengths of $AF$, $AE$, $AD$, $AC$, How can I calculate $BC$?
I tried to calculate $FD = sqrt{AF^2 - AD^2}$,
$DE = sqrt{AE^2 - AD^2}$,
$FE = DE + FD$,
$FC = AC - AF$.
But I don't know what I can do with all these lengths to compute $BC$. Any hints are appreciated.
geometry euclidean-geometry
geometry euclidean-geometry
asked Dec 29 '18 at 9:03
Mohamed MagdyMohamed Magdy
31127
asked Dec 29 '18 at 9:03
Mohamed MagdyMohamed Magdy
31127
asked Dec 29 '18 at 9:03
Mohamed MagdyMohamed Magdy
31127
asked Dec 29 '18 at 9:03
Mohamed MagdyMohamed Magdy
31127
asked Dec 29 '18 at 9:03
Mohamed MagdyMohamed Magdy
31127
31127
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $alpha= angle_{FAD}$, $beta= angle_{EAD}$.
Then you are given $cosalpha= AD/AF$, and similarly $cosbeta$.
You can now compute $cos (alpha+beta)$.
As $alpha+beta = angle_{CAB}$, you have the hypothenuse $AB$ of the right triangle $ABC$.
You can finish the calculation now.
answered Dec 29 '18 at 9:14
A. PongráczA. Pongrácz
6,1171929
$endgroup$
$begingroup$
$D$ isn't on an angle bisector in general.
$endgroup$
– jmerry
Dec 29 '18 at 9:16
$begingroup$
Yes, I realized it. Thank you!
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:16
add a comment |
$begingroup$
Trigonometry is the way to go.
We know everything about triangles $ADF$ and $ADE$. That means we know $angle FAD$ and $angle DAE$. From that, we know their sum, $angle FAE=angle CAB$. That's enough to figure out everything we need about the big triangle.
answered Dec 29 '18 at 9:15
jmerryjmerry
17k11633
$endgroup$
$begingroup$
Should I use similarity to figure out $AB$?
$endgroup$
– Mohamed Magdy
Dec 29 '18 at 9:20
$begingroup$
No, because the big triangle isn't similar to anything else in the picture. Use trigonometry; there will be an angle-sum formula in there.
$endgroup$
– jmerry
Dec 29 '18 at 9:21
$begingroup$
No @Mohamed because it is not said that some of triangles are similar.
$endgroup$
– user376343
Dec 29 '18 at 9:22
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $alpha= angle_{FAD}$, $beta= angle_{EAD}$.
Then you are given $cosalpha= AD/AF$, and similarly $cosbeta$.
You can now compute $cos (alpha+beta)$.
As $alpha+beta = angle_{CAB}$, you have the hypothenuse $AB$ of the right triangle $ABC$.
You can finish the calculation now.
answered Dec 29 '18 at 9:14
A. PongráczA. Pongrácz
6,1171929
$endgroup$
$begingroup$
$D$ isn't on an angle bisector in general.
$endgroup$
– jmerry
Dec 29 '18 at 9:16
$begingroup$
Yes, I realized it. Thank you!
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:16
add a comment |
$begingroup$
Let $alpha= angle_{FAD}$, $beta= angle_{EAD}$.
Then you are given $cosalpha= AD/AF$, and similarly $cosbeta$.
You can now compute $cos (alpha+beta)$.
As $alpha+beta = angle_{CAB}$, you have the hypothenuse $AB$ of the right triangle $ABC$.
You can finish the calculation now.
answered Dec 29 '18 at 9:14
A. PongráczA. Pongrácz
6,1171929
$endgroup$
$begingroup$
$D$ isn't on an angle bisector in general.
$endgroup$
– jmerry
Dec 29 '18 at 9:16
$begingroup$
Yes, I realized it. Thank you!
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:16
add a comment |
$begingroup$
Let $alpha= angle_{FAD}$, $beta= angle_{EAD}$.
Then you are given $cosalpha= AD/AF$, and similarly $cosbeta$.
You can now compute $cos (alpha+beta)$.
As $alpha+beta = angle_{CAB}$, you have the hypothenuse $AB$ of the right triangle $ABC$.
You can finish the calculation now.
answered Dec 29 '18 at 9:14
A. PongráczA. Pongrácz
6,1171929
$endgroup$
Let $alpha= angle_{FAD}$, $beta= angle_{EAD}$.
Then you are given $cosalpha= AD/AF$, and similarly $cosbeta$.
You can now compute $cos (alpha+beta)$.
As $alpha+beta = angle_{CAB}$, you have the hypothenuse $AB$ of the right triangle $ABC$.
You can finish the calculation now.
answered Dec 29 '18 at 9:14
A. PongráczA. Pongrácz
6,1171929
edited Dec 29 '18 at 9:16
answered Dec 29 '18 at 9:14
A. PongráczA. Pongrácz
6,1171929
answered Dec 29 '18 at 9:14
A. PongráczA. Pongrácz
6,1171929
answered Dec 29 '18 at 9:14
A. PongráczA. Pongrácz
6,1171929
6,1171929
$begingroup$
$D$ isn't on an angle bisector in general.
$endgroup$
– jmerry
Dec 29 '18 at 9:16
$begingroup$
Yes, I realized it. Thank you!
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:16
add a comment |
$begingroup$
$D$ isn't on an angle bisector in general.
$endgroup$
– jmerry
Dec 29 '18 at 9:16
$begingroup$
Yes, I realized it. Thank you!
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:16
$begingroup$
$D$ isn't on an angle bisector in general.
$endgroup$
– jmerry
Dec 29 '18 at 9:16
$begingroup$
$D$ isn't on an angle bisector in general.
$endgroup$
– jmerry
Dec 29 '18 at 9:16
$begingroup$
Yes, I realized it. Thank you!
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:16
$begingroup$
Yes, I realized it. Thank you!
$endgroup$
– A. Pongrácz
Dec 29 '18 at 9:16
add a comment |
$begingroup$
Trigonometry is the way to go.
We know everything about triangles $ADF$ and $ADE$. That means we know $angle FAD$ and $angle DAE$. From that, we know their sum, $angle FAE=angle CAB$. That's enough to figure out everything we need about the big triangle.
answered Dec 29 '18 at 9:15
jmerryjmerry
17k11633
$endgroup$
$begingroup$
Should I use similarity to figure out $AB$?
$endgroup$
– Mohamed Magdy
Dec 29 '18 at 9:20
$begingroup$
No, because the big triangle isn't similar to anything else in the picture. Use trigonometry; there will be an angle-sum formula in there.
$endgroup$
– jmerry
Dec 29 '18 at 9:21
$begingroup$
No @Mohamed because it is not said that some of triangles are similar.
$endgroup$
– user376343
Dec 29 '18 at 9:22
add a comment |
$begingroup$
Trigonometry is the way to go.
We know everything about triangles $ADF$ and $ADE$. That means we know $angle FAD$ and $angle DAE$. From that, we know their sum, $angle FAE=angle CAB$. That's enough to figure out everything we need about the big triangle.
answered Dec 29 '18 at 9:15
jmerryjmerry
17k11633
$endgroup$
$begingroup$
Should I use similarity to figure out $AB$?
$endgroup$
– Mohamed Magdy
Dec 29 '18 at 9:20
$begingroup$
No, because the big triangle isn't similar to anything else in the picture. Use trigonometry; there will be an angle-sum formula in there.
$endgroup$
– jmerry
Dec 29 '18 at 9:21
$begingroup$
No @Mohamed because it is not said that some of triangles are similar.
$endgroup$
– user376343
Dec 29 '18 at 9:22
add a comment |
$begingroup$
Trigonometry is the way to go.
We know everything about triangles $ADF$ and $ADE$. That means we know $angle FAD$ and $angle DAE$. From that, we know their sum, $angle FAE=angle CAB$. That's enough to figure out everything we need about the big triangle.
answered Dec 29 '18 at 9:15
jmerryjmerry
17k11633
$endgroup$
Trigonometry is the way to go.
We know everything about triangles $ADF$ and $ADE$. That means we know $angle FAD$ and $angle DAE$. From that, we know their sum, $angle FAE=angle CAB$. That's enough to figure out everything we need about the big triangle.
answered Dec 29 '18 at 9:15
jmerryjmerry
17k11633
answered Dec 29 '18 at 9:15
jmerryjmerry
17k11633
answered Dec 29 '18 at 9:15
jmerryjmerry
17k11633
answered Dec 29 '18 at 9:15
jmerryjmerry
17k11633
17k11633
$begingroup$
Should I use similarity to figure out $AB$?
$endgroup$
– Mohamed Magdy
Dec 29 '18 at 9:20
$begingroup$
No, because the big triangle isn't similar to anything else in the picture. Use trigonometry; there will be an angle-sum formula in there.
$endgroup$
– jmerry
Dec 29 '18 at 9:21
$begingroup$
No @Mohamed because it is not said that some of triangles are similar.
$endgroup$
– user376343
Dec 29 '18 at 9:22
add a comment |
$begingroup$
Should I use similarity to figure out $AB$?
$endgroup$
– Mohamed Magdy
Dec 29 '18 at 9:20
$begingroup$
No, because the big triangle isn't similar to anything else in the picture. Use trigonometry; there will be an angle-sum formula in there.
$endgroup$
– jmerry
Dec 29 '18 at 9:21
$begingroup$
No @Mohamed because it is not said that some of triangles are similar.
$endgroup$
– user376343
Dec 29 '18 at 9:22
$begingroup$
Should I use similarity to figure out $AB$?
$endgroup$
– Mohamed Magdy
Dec 29 '18 at 9:20
$begingroup$
Should I use similarity to figure out $AB$?
$endgroup$
– Mohamed Magdy
Dec 29 '18 at 9:20
$begingroup$
No, because the big triangle isn't similar to anything else in the picture. Use trigonometry; there will be an angle-sum formula in there.
$endgroup$
– jmerry
Dec 29 '18 at 9:21
$begingroup$
No, because the big triangle isn't similar to anything else in the picture. Use trigonometry; there will be an angle-sum formula in there.
$endgroup$
– jmerry
Dec 29 '18 at 9:21
$begingroup$
No @Mohamed because it is not said that some of triangles are similar.
$endgroup$
– user376343
Dec 29 '18 at 9:22
$begingroup$
No @Mohamed because it is not said that some of triangles are similar.
$endgroup$
– user376343
Dec 29 '18 at 9:22
add a comment |
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