Mixing
Geometry with circle and triangle. [closed]
$begingroup$
Any other solutions(advice) are welcome.
What I am asking is this because I am studying mathematics through feedback process of my solution and learning new solutions.
Please, release Hold on.
When will the hold be resolved? I want to see your new solutions(or advices).
$overline{AE}$ is a diameter.
and $overline{AB}=1,;;; overline{AC}=2$
and $overline{BD}:overline{CD}=3:2$
Find the area of $triangle ABC$ .
geometry proof-verification
asked Jan 23 at 7:01
mina_worldmina_world
1799
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closed as unclear what you're asking by Riccardo.Alestra, Namaste, Jyrki Lahtonen, Adrian Keister, Lee David Chung Lin Jan 23 at 14:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Any other solutions(advice) are welcome.
What I am asking is this because I am studying mathematics through feedback process of my solution and learning new solutions.
Please, release Hold on.
When will the hold be resolved? I want to see your new solutions(or advices).
$overline{AE}$ is a diameter.
and $overline{AB}=1,;;; overline{AC}=2$
and $overline{BD}:overline{CD}=3:2$
Find the area of $triangle ABC$ .
geometry proof-verification
asked Jan 23 at 7:01
mina_worldmina_world
1799
$endgroup$
closed as unclear what you're asking by Riccardo.Alestra, Namaste, Jyrki Lahtonen, Adrian Keister, Lee David Chung Lin Jan 23 at 14:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Just to be clear: the question is asking for alternative solutions to your problem? If you're trying to share your solution by answering your own question there is an option to do so when asking a question.
$endgroup$
– user574848
Jan 23 at 7:11
$begingroup$
@user574848 Yes, it's good option. Thank you for telling me.
$endgroup$
– mina_world
Jan 23 at 7:22
add a comment |
$begingroup$
Any other solutions(advice) are welcome.
What I am asking is this because I am studying mathematics through feedback process of my solution and learning new solutions.
Please, release Hold on.
When will the hold be resolved? I want to see your new solutions(or advices).
$overline{AE}$ is a diameter.
and $overline{AB}=1,;;; overline{AC}=2$
and $overline{BD}:overline{CD}=3:2$
Find the area of $triangle ABC$ .
geometry proof-verification
asked Jan 23 at 7:01
mina_worldmina_world
1799
$endgroup$
Any other solutions(advice) are welcome.
What I am asking is this because I am studying mathematics through feedback process of my solution and learning new solutions.
Please, release Hold on.
When will the hold be resolved? I want to see your new solutions(or advices).
$overline{AE}$ is a diameter.
and $overline{AB}=1,;;; overline{AC}=2$
and $overline{BD}:overline{CD}=3:2$
Find the area of $triangle ABC$ .
geometry proof-verification
geometry proof-verification
asked Jan 23 at 7:01
mina_worldmina_world
1799
asked Jan 23 at 7:01
mina_worldmina_world
1799
edited Jan 25 at 7:28
mina_world
asked Jan 23 at 7:01
mina_worldmina_world
1799
asked Jan 23 at 7:01
mina_worldmina_world
1799
asked Jan 23 at 7:01
mina_worldmina_world
1799
1799
closed as unclear what you're asking by Riccardo.Alestra, Namaste, Jyrki Lahtonen, Adrian Keister, Lee David Chung Lin Jan 23 at 14:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Riccardo.Alestra, Namaste, Jyrki Lahtonen, Adrian Keister, Lee David Chung Lin Jan 23 at 14:50
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Just to be clear: the question is asking for alternative solutions to your problem? If you're trying to share your solution by answering your own question there is an option to do so when asking a question.
$endgroup$
– user574848
Jan 23 at 7:11
$begingroup$
@user574848 Yes, it's good option. Thank you for telling me.
$endgroup$
– mina_world
Jan 23 at 7:22
add a comment |
1
$begingroup$
Just to be clear: the question is asking for alternative solutions to your problem? If you're trying to share your solution by answering your own question there is an option to do so when asking a question.
$endgroup$
– user574848
Jan 23 at 7:11
$begingroup$
@user574848 Yes, it's good option. Thank you for telling me.
$endgroup$
– mina_world
Jan 23 at 7:22
1
1
$begingroup$
Just to be clear: the question is asking for alternative solutions to your problem? If you're trying to share your solution by answering your own question there is an option to do so when asking a question.
$endgroup$
– user574848
Jan 23 at 7:11
$begingroup$
Just to be clear: the question is asking for alternative solutions to your problem? If you're trying to share your solution by answering your own question there is an option to do so when asking a question.
$endgroup$
– user574848
Jan 23 at 7:11
$begingroup$
@user574848 Yes, it's good option. Thank you for telling me.
$endgroup$
– mina_world
Jan 23 at 7:22
$begingroup$
@user574848 Yes, it's good option. Thank you for telling me.
$endgroup$
– mina_world
Jan 23 at 7:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a completely different soluiton based on trigonometry.
From triangle $ABD$:
$$frac{3x}{sinalpha}=frac{l}{sin(90^circ-beta)}tag{1}$$
From triangle $ADC$:
$$frac{2x}{sinbeta}=frac{l}{sin(90^circ-alpha)}tag{2}$$
From triangle $ABC$:
$$frac{1}{sin(90^circ-alpha)}=frac{2}{sin(90^circ-beta)}tag{3}$$
Rewrite (1),(2)(3):
$$frac{3x}{sinalpha}=frac{l}{cosbeta}tag{4}$$
$$frac{2x}{sinbeta}=frac{l}{cosalpha}tag{5}$$
$$cosbeta=2cosalphatag{6}$$
We don't care about $l$, just divide (4) by (5) to eliminate it:
$$frac{3sinbeta}{2sinalpha}=frac{cosalpha}{cosbeta}$$
$$3sinbetacosbeta=2sinalphacosalphatag{7}$$
Now you have (6) and (7), two equations with two unknown angles. Divide (7) by (6) and you get:
$$3sinbeta=sinalphatag{8}$$
Now divide (6) by 2, square and add to (8) squared to eliminate $alpha$:
$$9sin^2beta+frac14cos^2beta=1$$
$$36sin^2beta+cos^2beta=4$$
$$35sin^2beta=3$$
The rest is straightforward:
$$sinbeta=frac{sqrt{3}}{sqrt{35}}, cosbeta=frac{4sqrt{2}}{sqrt{35}}$$
From (8):
$$sinalpha=frac{3sqrt{3}}{sqrt{35}}, cosalpha=frac{2sqrt{2}}{sqrt{35}}$$
Now:
$$S=frac12 ABcdot h=frac12 cdot 1 cdot 2 sin(alpha+beta)=sin(alpha+beta)$$
$$S=sinalphacosbeta+cosalphasinbeta=frac{2sqrt{6}}{5}$$
answered Jan 23 at 12:49
OldboyOldboy
8,66211036
$endgroup$
$begingroup$
Yes, good idea. In fact, (8)$;3sin;beta=sin;alpha;$ $Rightarrow$$;; 3times frac{overline{EC}}{overline{AE}}=frac{overline{EB}}{overline{AE}} $$;;Rightarrow ;3overline{EC}=overline{EB}$ Thank you very much.
$endgroup$
– mina_world
Jan 23 at 13:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a completely different soluiton based on trigonometry.
From triangle $ABD$:
$$frac{3x}{sinalpha}=frac{l}{sin(90^circ-beta)}tag{1}$$
From triangle $ADC$:
$$frac{2x}{sinbeta}=frac{l}{sin(90^circ-alpha)}tag{2}$$
From triangle $ABC$:
$$frac{1}{sin(90^circ-alpha)}=frac{2}{sin(90^circ-beta)}tag{3}$$
Rewrite (1),(2)(3):
$$frac{3x}{sinalpha}=frac{l}{cosbeta}tag{4}$$
$$frac{2x}{sinbeta}=frac{l}{cosalpha}tag{5}$$
$$cosbeta=2cosalphatag{6}$$
We don't care about $l$, just divide (4) by (5) to eliminate it:
$$frac{3sinbeta}{2sinalpha}=frac{cosalpha}{cosbeta}$$
$$3sinbetacosbeta=2sinalphacosalphatag{7}$$
Now you have (6) and (7), two equations with two unknown angles. Divide (7) by (6) and you get:
$$3sinbeta=sinalphatag{8}$$
Now divide (6) by 2, square and add to (8) squared to eliminate $alpha$:
$$9sin^2beta+frac14cos^2beta=1$$
$$36sin^2beta+cos^2beta=4$$
$$35sin^2beta=3$$
The rest is straightforward:
$$sinbeta=frac{sqrt{3}}{sqrt{35}}, cosbeta=frac{4sqrt{2}}{sqrt{35}}$$
From (8):
$$sinalpha=frac{3sqrt{3}}{sqrt{35}}, cosalpha=frac{2sqrt{2}}{sqrt{35}}$$
Now:
$$S=frac12 ABcdot h=frac12 cdot 1 cdot 2 sin(alpha+beta)=sin(alpha+beta)$$
$$S=sinalphacosbeta+cosalphasinbeta=frac{2sqrt{6}}{5}$$
answered Jan 23 at 12:49
OldboyOldboy
8,66211036
$endgroup$
$begingroup$
Yes, good idea. In fact, (8)$;3sin;beta=sin;alpha;$ $Rightarrow$$;; 3times frac{overline{EC}}{overline{AE}}=frac{overline{EB}}{overline{AE}} $$;;Rightarrow ;3overline{EC}=overline{EB}$ Thank you very much.
$endgroup$
– mina_world
Jan 23 at 13:35
add a comment |
$begingroup$
Here is a completely different soluiton based on trigonometry.
From triangle $ABD$:
$$frac{3x}{sinalpha}=frac{l}{sin(90^circ-beta)}tag{1}$$
From triangle $ADC$:
$$frac{2x}{sinbeta}=frac{l}{sin(90^circ-alpha)}tag{2}$$
From triangle $ABC$:
$$frac{1}{sin(90^circ-alpha)}=frac{2}{sin(90^circ-beta)}tag{3}$$
Rewrite (1),(2)(3):
$$frac{3x}{sinalpha}=frac{l}{cosbeta}tag{4}$$
$$frac{2x}{sinbeta}=frac{l}{cosalpha}tag{5}$$
$$cosbeta=2cosalphatag{6}$$
We don't care about $l$, just divide (4) by (5) to eliminate it:
$$frac{3sinbeta}{2sinalpha}=frac{cosalpha}{cosbeta}$$
$$3sinbetacosbeta=2sinalphacosalphatag{7}$$
Now you have (6) and (7), two equations with two unknown angles. Divide (7) by (6) and you get:
$$3sinbeta=sinalphatag{8}$$
Now divide (6) by 2, square and add to (8) squared to eliminate $alpha$:
$$9sin^2beta+frac14cos^2beta=1$$
$$36sin^2beta+cos^2beta=4$$
$$35sin^2beta=3$$
The rest is straightforward:
$$sinbeta=frac{sqrt{3}}{sqrt{35}}, cosbeta=frac{4sqrt{2}}{sqrt{35}}$$
From (8):
$$sinalpha=frac{3sqrt{3}}{sqrt{35}}, cosalpha=frac{2sqrt{2}}{sqrt{35}}$$
Now:
$$S=frac12 ABcdot h=frac12 cdot 1 cdot 2 sin(alpha+beta)=sin(alpha+beta)$$
$$S=sinalphacosbeta+cosalphasinbeta=frac{2sqrt{6}}{5}$$
answered Jan 23 at 12:49
OldboyOldboy
8,66211036
$endgroup$
$begingroup$
Yes, good idea. In fact, (8)$;3sin;beta=sin;alpha;$ $Rightarrow$$;; 3times frac{overline{EC}}{overline{AE}}=frac{overline{EB}}{overline{AE}} $$;;Rightarrow ;3overline{EC}=overline{EB}$ Thank you very much.
$endgroup$
– mina_world
Jan 23 at 13:35
add a comment |
$begingroup$
Here is a completely different soluiton based on trigonometry.
From triangle $ABD$:
$$frac{3x}{sinalpha}=frac{l}{sin(90^circ-beta)}tag{1}$$
From triangle $ADC$:
$$frac{2x}{sinbeta}=frac{l}{sin(90^circ-alpha)}tag{2}$$
From triangle $ABC$:
$$frac{1}{sin(90^circ-alpha)}=frac{2}{sin(90^circ-beta)}tag{3}$$
Rewrite (1),(2)(3):
$$frac{3x}{sinalpha}=frac{l}{cosbeta}tag{4}$$
$$frac{2x}{sinbeta}=frac{l}{cosalpha}tag{5}$$
$$cosbeta=2cosalphatag{6}$$
We don't care about $l$, just divide (4) by (5) to eliminate it:
$$frac{3sinbeta}{2sinalpha}=frac{cosalpha}{cosbeta}$$
$$3sinbetacosbeta=2sinalphacosalphatag{7}$$
Now you have (6) and (7), two equations with two unknown angles. Divide (7) by (6) and you get:
$$3sinbeta=sinalphatag{8}$$
Now divide (6) by 2, square and add to (8) squared to eliminate $alpha$:
$$9sin^2beta+frac14cos^2beta=1$$
$$36sin^2beta+cos^2beta=4$$
$$35sin^2beta=3$$
The rest is straightforward:
$$sinbeta=frac{sqrt{3}}{sqrt{35}}, cosbeta=frac{4sqrt{2}}{sqrt{35}}$$
From (8):
$$sinalpha=frac{3sqrt{3}}{sqrt{35}}, cosalpha=frac{2sqrt{2}}{sqrt{35}}$$
Now:
$$S=frac12 ABcdot h=frac12 cdot 1 cdot 2 sin(alpha+beta)=sin(alpha+beta)$$
$$S=sinalphacosbeta+cosalphasinbeta=frac{2sqrt{6}}{5}$$
answered Jan 23 at 12:49
OldboyOldboy
8,66211036
$endgroup$
Here is a completely different soluiton based on trigonometry.
From triangle $ABD$:
$$frac{3x}{sinalpha}=frac{l}{sin(90^circ-beta)}tag{1}$$
From triangle $ADC$:
$$frac{2x}{sinbeta}=frac{l}{sin(90^circ-alpha)}tag{2}$$
From triangle $ABC$:
$$frac{1}{sin(90^circ-alpha)}=frac{2}{sin(90^circ-beta)}tag{3}$$
Rewrite (1),(2)(3):
$$frac{3x}{sinalpha}=frac{l}{cosbeta}tag{4}$$
$$frac{2x}{sinbeta}=frac{l}{cosalpha}tag{5}$$
$$cosbeta=2cosalphatag{6}$$
We don't care about $l$, just divide (4) by (5) to eliminate it:
$$frac{3sinbeta}{2sinalpha}=frac{cosalpha}{cosbeta}$$
$$3sinbetacosbeta=2sinalphacosalphatag{7}$$
Now you have (6) and (7), two equations with two unknown angles. Divide (7) by (6) and you get:
$$3sinbeta=sinalphatag{8}$$
Now divide (6) by 2, square and add to (8) squared to eliminate $alpha$:
$$9sin^2beta+frac14cos^2beta=1$$
$$36sin^2beta+cos^2beta=4$$
$$35sin^2beta=3$$
The rest is straightforward:
$$sinbeta=frac{sqrt{3}}{sqrt{35}}, cosbeta=frac{4sqrt{2}}{sqrt{35}}$$
From (8):
$$sinalpha=frac{3sqrt{3}}{sqrt{35}}, cosalpha=frac{2sqrt{2}}{sqrt{35}}$$
Now:
$$S=frac12 ABcdot h=frac12 cdot 1 cdot 2 sin(alpha+beta)=sin(alpha+beta)$$
$$S=sinalphacosbeta+cosalphasinbeta=frac{2sqrt{6}}{5}$$
answered Jan 23 at 12:49
OldboyOldboy
8,66211036
answered Jan 23 at 12:49
OldboyOldboy
8,66211036
answered Jan 23 at 12:49
OldboyOldboy
8,66211036
answered Jan 23 at 12:49
OldboyOldboy
8,66211036
8,66211036
$begingroup$
Yes, good idea. In fact, (8)$;3sin;beta=sin;alpha;$ $Rightarrow$$;; 3times frac{overline{EC}}{overline{AE}}=frac{overline{EB}}{overline{AE}} $$;;Rightarrow ;3overline{EC}=overline{EB}$ Thank you very much.
$endgroup$
– mina_world
Jan 23 at 13:35
add a comment |
$begingroup$
Yes, good idea. In fact, (8)$;3sin;beta=sin;alpha;$ $Rightarrow$$;; 3times frac{overline{EC}}{overline{AE}}=frac{overline{EB}}{overline{AE}} $$;;Rightarrow ;3overline{EC}=overline{EB}$ Thank you very much.
$endgroup$
– mina_world
Jan 23 at 13:35
$begingroup$
Yes, good idea. In fact, (8)$;3sin;beta=sin;alpha;$ $Rightarrow$$;; 3times frac{overline{EC}}{overline{AE}}=frac{overline{EB}}{overline{AE}} $$;;Rightarrow ;3overline{EC}=overline{EB}$ Thank you very much.
$endgroup$
– mina_world
Jan 23 at 13:35
$begingroup$
Yes, good idea. In fact, (8)$;3sin;beta=sin;alpha;$ $Rightarrow$$;; 3times frac{overline{EC}}{overline{AE}}=frac{overline{EB}}{overline{AE}} $$;;Rightarrow ;3overline{EC}=overline{EB}$ Thank you very much.
$endgroup$
– mina_world
Jan 23 at 13:35
add a comment |
1
$begingroup$
Just to be clear: the question is asking for alternative solutions to your problem? If you're trying to share your solution by answering your own question there is an option to do so when asking a question.
$endgroup$
– user574848
Jan 23 at 7:11
$begingroup$
@user574848 Yes, it's good option. Thank you for telling me.
$endgroup$
– mina_world
Jan 23 at 7:22