How to make correct system of equations to solve for the angles in this triangle?
I'm trying to solve this triangle for $X$. Thereby, I've tried to make correct system of equations. What would be the correct equations?
Here are the equations I can find
- In $triangle ABC$, recalling that $angle ACD = y$
$$48 + 24 + x + 12 + y = 180$$
- In $triangle ADC$
$$84+x + y = 180$$
However, I'm getting the same equations.
Regards
triangle
asked Dec 1 at 10:09
Hamilton
1848
add a comment |
I'm trying to solve this triangle for $X$. Thereby, I've tried to make correct system of equations. What would be the correct equations?
Here are the equations I can find
- In $triangle ABC$, recalling that $angle ACD = y$
$$48 + 24 + x + 12 + y = 180$$
- In $triangle ADC$
$$84+x + y = 180$$
However, I'm getting the same equations.
Regards
triangle
asked Dec 1 at 10:09
Hamilton
1848
Which equations can you find?
– Arthur
Dec 1 at 10:34
@Arthur See the update, please.
– Hamilton
Dec 1 at 10:57
Geogebra suggests the $x=80$.
– Michael Hoppe
Dec 1 at 11:44
1
No, the information is sufficient: just try to construct.
– Michael Hoppe
Dec 1 at 12:10
I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
– Fareed AF
Dec 8 at 14:13
add a comment |
I'm trying to solve this triangle for $X$. Thereby, I've tried to make correct system of equations. What would be the correct equations?
Here are the equations I can find
- In $triangle ABC$, recalling that $angle ACD = y$
$$48 + 24 + x + 12 + y = 180$$
- In $triangle ADC$
$$84+x + y = 180$$
However, I'm getting the same equations.
Regards
triangle
asked Dec 1 at 10:09
Hamilton
1848
I'm trying to solve this triangle for $X$. Thereby, I've tried to make correct system of equations. What would be the correct equations?
Here are the equations I can find
- In $triangle ABC$, recalling that $angle ACD = y$
$$48 + 24 + x + 12 + y = 180$$
- In $triangle ADC$
$$84+x + y = 180$$
However, I'm getting the same equations.
Regards
triangle
triangle
asked Dec 1 at 10:09
Hamilton
1848
asked Dec 1 at 10:09
Hamilton
1848
edited Dec 1 at 10:57
asked Dec 1 at 10:09
Hamilton
1848
asked Dec 1 at 10:09
Hamilton
1848
asked Dec 1 at 10:09
Hamilton
1848
1848
Which equations can you find?
– Arthur
Dec 1 at 10:34
@Arthur See the update, please.
– Hamilton
Dec 1 at 10:57
Geogebra suggests the $x=80$.
– Michael Hoppe
Dec 1 at 11:44
1
No, the information is sufficient: just try to construct.
– Michael Hoppe
Dec 1 at 12:10
I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
– Fareed AF
Dec 8 at 14:13
add a comment |
Which equations can you find?
– Arthur
Dec 1 at 10:34
@Arthur See the update, please.
– Hamilton
Dec 1 at 10:57
Geogebra suggests the $x=80$.
– Michael Hoppe
Dec 1 at 11:44
1
No, the information is sufficient: just try to construct.
– Michael Hoppe
Dec 1 at 12:10
I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
– Fareed AF
Dec 8 at 14:13
Which equations can you find?
– Arthur
Dec 1 at 10:34
Which equations can you find?
– Arthur
Dec 1 at 10:34
@Arthur See the update, please.
– Hamilton
Dec 1 at 10:57
@Arthur See the update, please.
– Hamilton
Dec 1 at 10:57
Geogebra suggests the $x=80$.
– Michael Hoppe
Dec 1 at 11:44
Geogebra suggests the $x=80$.
– Michael Hoppe
Dec 1 at 11:44
1
1
No, the information is sufficient: just try to construct.
– Michael Hoppe
Dec 1 at 12:10
No, the information is sufficient: just try to construct.
– Michael Hoppe
Dec 1 at 12:10
I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
– Fareed AF
Dec 8 at 14:13
I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
– Fareed AF
Dec 8 at 14:13
add a comment |
2 Answers
2
active
oldest
votes
The key insight is to observe that $angle(BDA)=angle(BDC)=138$. From here in triangle $BDC$ we have $AB/sin(138)=BD/sin(24)$ and from triangle BCD we get
$BC/sin(138)=BD/sin(12)$. Divingig both equations gives
$$BC=frac{ABsin(24)}{sin(12)}.$$
Now use law of cosine in $ABC$ to get
$$AC^2=AB^2+BC^2-2ABcdot BCcos(48)
=BC^2left(frac{sin^2(12)}{sin^2(24)}+1
-2frac{sin(12)}{sin(24)}{cos(48)}right).$$
Finally use law of sine in triangle $ABC$:
$$frac{sin(x+24)}{BC}=frac{sin(48)}{AC}$$
plug in $AB$ to get
$$sin(x+48)=frac{sin(48)}{sqrt{dfrac{sin^2(12)}{sin^2(24)}+1
-2dfrac{sin(12)}{sin(24)}{cos(48)}}}$$
answered Dec 1 at 14:59
Michael Hoppe
10.8k31834
add a comment |
I use degrees, but omit the notation.
With the use of the law of sines we have $$frac{AD}{sin 18}=frac{AB}{sin 138}=frac{BD}{sin 24}.$$
One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$
Similarly we obtain the sides of $triangle BCD.$
It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=frac{CDsin 84}{sin X}=frac{BC sin 48}{sin(24+X)}$$ From this one gets easily $tan X.$
answered Dec 1 at 11:02
user376343
2,7782822
Why did you use law of sines?
– Hamilton
Dec 1 at 11:06
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
The key insight is to observe that $angle(BDA)=angle(BDC)=138$. From here in triangle $BDC$ we have $AB/sin(138)=BD/sin(24)$ and from triangle BCD we get
$BC/sin(138)=BD/sin(12)$. Divingig both equations gives
$$BC=frac{ABsin(24)}{sin(12)}.$$
Now use law of cosine in $ABC$ to get
$$AC^2=AB^2+BC^2-2ABcdot BCcos(48)
=BC^2left(frac{sin^2(12)}{sin^2(24)}+1
-2frac{sin(12)}{sin(24)}{cos(48)}right).$$
Finally use law of sine in triangle $ABC$:
$$frac{sin(x+24)}{BC}=frac{sin(48)}{AC}$$
plug in $AB$ to get
$$sin(x+48)=frac{sin(48)}{sqrt{dfrac{sin^2(12)}{sin^2(24)}+1
-2dfrac{sin(12)}{sin(24)}{cos(48)}}}$$
answered Dec 1 at 14:59
Michael Hoppe
10.8k31834
add a comment |
The key insight is to observe that $angle(BDA)=angle(BDC)=138$. From here in triangle $BDC$ we have $AB/sin(138)=BD/sin(24)$ and from triangle BCD we get
$BC/sin(138)=BD/sin(12)$. Divingig both equations gives
$$BC=frac{ABsin(24)}{sin(12)}.$$
Now use law of cosine in $ABC$ to get
$$AC^2=AB^2+BC^2-2ABcdot BCcos(48)
=BC^2left(frac{sin^2(12)}{sin^2(24)}+1
-2frac{sin(12)}{sin(24)}{cos(48)}right).$$
Finally use law of sine in triangle $ABC$:
$$frac{sin(x+24)}{BC}=frac{sin(48)}{AC}$$
plug in $AB$ to get
$$sin(x+48)=frac{sin(48)}{sqrt{dfrac{sin^2(12)}{sin^2(24)}+1
-2dfrac{sin(12)}{sin(24)}{cos(48)}}}$$
answered Dec 1 at 14:59
Michael Hoppe
10.8k31834
add a comment |
The key insight is to observe that $angle(BDA)=angle(BDC)=138$. From here in triangle $BDC$ we have $AB/sin(138)=BD/sin(24)$ and from triangle BCD we get
$BC/sin(138)=BD/sin(12)$. Divingig both equations gives
$$BC=frac{ABsin(24)}{sin(12)}.$$
Now use law of cosine in $ABC$ to get
$$AC^2=AB^2+BC^2-2ABcdot BCcos(48)
=BC^2left(frac{sin^2(12)}{sin^2(24)}+1
-2frac{sin(12)}{sin(24)}{cos(48)}right).$$
Finally use law of sine in triangle $ABC$:
$$frac{sin(x+24)}{BC}=frac{sin(48)}{AC}$$
plug in $AB$ to get
$$sin(x+48)=frac{sin(48)}{sqrt{dfrac{sin^2(12)}{sin^2(24)}+1
-2dfrac{sin(12)}{sin(24)}{cos(48)}}}$$
answered Dec 1 at 14:59
Michael Hoppe
10.8k31834
The key insight is to observe that $angle(BDA)=angle(BDC)=138$. From here in triangle $BDC$ we have $AB/sin(138)=BD/sin(24)$ and from triangle BCD we get
$BC/sin(138)=BD/sin(12)$. Divingig both equations gives
$$BC=frac{ABsin(24)}{sin(12)}.$$
Now use law of cosine in $ABC$ to get
$$AC^2=AB^2+BC^2-2ABcdot BCcos(48)
=BC^2left(frac{sin^2(12)}{sin^2(24)}+1
-2frac{sin(12)}{sin(24)}{cos(48)}right).$$
Finally use law of sine in triangle $ABC$:
$$frac{sin(x+24)}{BC}=frac{sin(48)}{AC}$$
plug in $AB$ to get
$$sin(x+48)=frac{sin(48)}{sqrt{dfrac{sin^2(12)}{sin^2(24)}+1
-2dfrac{sin(12)}{sin(24)}{cos(48)}}}$$
answered Dec 1 at 14:59
Michael Hoppe
10.8k31834
edited Dec 1 at 16:01
answered Dec 1 at 14:59
Michael Hoppe
10.8k31834
answered Dec 1 at 14:59
Michael Hoppe
10.8k31834
answered Dec 1 at 14:59
Michael Hoppe
10.8k31834
10.8k31834
add a comment |
add a comment |
I use degrees, but omit the notation.
With the use of the law of sines we have $$frac{AD}{sin 18}=frac{AB}{sin 138}=frac{BD}{sin 24}.$$
One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$
Similarly we obtain the sides of $triangle BCD.$
It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=frac{CDsin 84}{sin X}=frac{BC sin 48}{sin(24+X)}$$ From this one gets easily $tan X.$
answered Dec 1 at 11:02
user376343
2,7782822
Why did you use law of sines?
– Hamilton
Dec 1 at 11:06
add a comment |
I use degrees, but omit the notation.
With the use of the law of sines we have $$frac{AD}{sin 18}=frac{AB}{sin 138}=frac{BD}{sin 24}.$$
One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$
Similarly we obtain the sides of $triangle BCD.$
It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=frac{CDsin 84}{sin X}=frac{BC sin 48}{sin(24+X)}$$ From this one gets easily $tan X.$
answered Dec 1 at 11:02
user376343
2,7782822
Why did you use law of sines?
– Hamilton
Dec 1 at 11:06
add a comment |
I use degrees, but omit the notation.
With the use of the law of sines we have $$frac{AD}{sin 18}=frac{AB}{sin 138}=frac{BD}{sin 24}.$$
One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$
Similarly we obtain the sides of $triangle BCD.$
It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=frac{CDsin 84}{sin X}=frac{BC sin 48}{sin(24+X)}$$ From this one gets easily $tan X.$
answered Dec 1 at 11:02
user376343
2,7782822
I use degrees, but omit the notation.
With the use of the law of sines we have $$frac{AD}{sin 18}=frac{AB}{sin 138}=frac{BD}{sin 24}.$$
One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$
Similarly we obtain the sides of $triangle BCD.$
It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=frac{CDsin 84}{sin X}=frac{BC sin 48}{sin(24+X)}$$ From this one gets easily $tan X.$
answered Dec 1 at 11:02
user376343
2,7782822
answered Dec 1 at 11:02
user376343
2,7782822
answered Dec 1 at 11:02
user376343
2,7782822
answered Dec 1 at 11:02
user376343
2,7782822
2,7782822
Why did you use law of sines?
– Hamilton
Dec 1 at 11:06
add a comment |
Why did you use law of sines?
– Hamilton
Dec 1 at 11:06
Why did you use law of sines?
– Hamilton
Dec 1 at 11:06
Why did you use law of sines?
– Hamilton
Dec 1 at 11:06
add a comment |
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Which equations can you find?
– Arthur
Dec 1 at 10:34
@Arthur See the update, please.
– Hamilton
Dec 1 at 10:57
Geogebra suggests the $x=80$.
– Michael Hoppe
Dec 1 at 11:44
1
No, the information is sufficient: just try to construct.
– Michael Hoppe
Dec 1 at 12:10
I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
– Fareed AF
Dec 8 at 14:13