Can we find angles of the new triangle?
$begingroup$
In an equilateral triangle $Delta ABC$ with side $a$, an interior point $D$ is chosen and joined with vertices to form line segments $AD$,$BD$ and $CD$.
Also we know that $angle ADC=x$, $angle BDC =y$ and $angle ADB=z$
Now if a new triangle with line segments $AD$, $BD$ and $CD$ are formed, Is it possible to find the angles of this new triangle?
My try:
Let $AD=p$, $BD=q$ and $CD=r$
By cosine rule we have:
$$cos z=frac{p^2+q^2-a^2}{2pq}$$
$$cos x=frac{p^2+r^2-a^2}{2pr}$$
$$cos y=frac{r^2+q^2-a^2}{2rq}$$
which are three equations in three unknowns $p,q,r$.
Can we solve these?
geometry trigonometry triangle
asked Jan 25 at 15:52
Umesh shankarUmesh shankar
2,99331220
$endgroup$
add a comment |
$begingroup$
In an equilateral triangle $Delta ABC$ with side $a$, an interior point $D$ is chosen and joined with vertices to form line segments $AD$,$BD$ and $CD$.
Also we know that $angle ADC=x$, $angle BDC =y$ and $angle ADB=z$
Now if a new triangle with line segments $AD$, $BD$ and $CD$ are formed, Is it possible to find the angles of this new triangle?
My try:
Let $AD=p$, $BD=q$ and $CD=r$
By cosine rule we have:
$$cos z=frac{p^2+q^2-a^2}{2pq}$$
$$cos x=frac{p^2+r^2-a^2}{2pr}$$
$$cos y=frac{r^2+q^2-a^2}{2rq}$$
which are three equations in three unknowns $p,q,r$.
Can we solve these?
geometry trigonometry triangle
asked Jan 25 at 15:52
Umesh shankarUmesh shankar
2,99331220
$endgroup$
1
$begingroup$
Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
$endgroup$
– Ross Millikan
Jan 25 at 15:57
$begingroup$
But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
$endgroup$
– Umesh shankar
Jan 25 at 16:02
add a comment |
$begingroup$
In an equilateral triangle $Delta ABC$ with side $a$, an interior point $D$ is chosen and joined with vertices to form line segments $AD$,$BD$ and $CD$.
Also we know that $angle ADC=x$, $angle BDC =y$ and $angle ADB=z$
Now if a new triangle with line segments $AD$, $BD$ and $CD$ are formed, Is it possible to find the angles of this new triangle?
My try:
Let $AD=p$, $BD=q$ and $CD=r$
By cosine rule we have:
$$cos z=frac{p^2+q^2-a^2}{2pq}$$
$$cos x=frac{p^2+r^2-a^2}{2pr}$$
$$cos y=frac{r^2+q^2-a^2}{2rq}$$
which are three equations in three unknowns $p,q,r$.
Can we solve these?
geometry trigonometry triangle
asked Jan 25 at 15:52
Umesh shankarUmesh shankar
2,99331220
$endgroup$
In an equilateral triangle $Delta ABC$ with side $a$, an interior point $D$ is chosen and joined with vertices to form line segments $AD$,$BD$ and $CD$.
Also we know that $angle ADC=x$, $angle BDC =y$ and $angle ADB=z$
Now if a new triangle with line segments $AD$, $BD$ and $CD$ are formed, Is it possible to find the angles of this new triangle?
My try:
Let $AD=p$, $BD=q$ and $CD=r$
By cosine rule we have:
$$cos z=frac{p^2+q^2-a^2}{2pq}$$
$$cos x=frac{p^2+r^2-a^2}{2pr}$$
$$cos y=frac{r^2+q^2-a^2}{2rq}$$
which are three equations in three unknowns $p,q,r$.
Can we solve these?
geometry trigonometry triangle
geometry trigonometry triangle
asked Jan 25 at 15:52
Umesh shankarUmesh shankar
2,99331220
asked Jan 25 at 15:52
Umesh shankarUmesh shankar
2,99331220
asked Jan 25 at 15:52
Umesh shankarUmesh shankar
2,99331220
asked Jan 25 at 15:52
Umesh shankarUmesh shankar
2,99331220
asked Jan 25 at 15:52
Umesh shankarUmesh shankar
2,99331220
2,99331220
1
$begingroup$
Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
$endgroup$
– Ross Millikan
Jan 25 at 15:57
$begingroup$
But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
$endgroup$
– Umesh shankar
Jan 25 at 16:02
add a comment |
1
$begingroup$
Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
$endgroup$
– Ross Millikan
Jan 25 at 15:57
$begingroup$
But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
$endgroup$
– Umesh shankar
Jan 25 at 16:02
1
1
$begingroup$
Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
$endgroup$
– Ross Millikan
Jan 25 at 15:57
$begingroup$
Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
$endgroup$
– Ross Millikan
Jan 25 at 15:57
$begingroup$
But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
$endgroup$
– Umesh shankar
Jan 25 at 16:02
$begingroup$
But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
$endgroup$
– Umesh shankar
Jan 25 at 16:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.
Let $R$ denote a rotation by $frac{pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $triangle CDD'cong T$. As rotation preserves angles too, $angle BCD'=angle BR(A)R(D)=angle BAD$. Therefore, $angle DCD'=angle DCB+angle BAD= 2pi-theta_A-theta_C-frac{pi}{3}=theta_B-frac{pi}{3}$. Therefore, angle opposite to $BD$ in $T = theta_B-frac{pi}{3}$.
Therefore, by symmetry, the angles of $T$ are $theta_A-frac{pi}{3}, theta_B-frac{pi}{3}$ and $theta_C-frac{pi}{3}$.
$blacksquare$
answered Jan 25 at 16:49
Anubhab GhosalAnubhab Ghosal
1,22319
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.
Let $R$ denote a rotation by $frac{pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $triangle CDD'cong T$. As rotation preserves angles too, $angle BCD'=angle BR(A)R(D)=angle BAD$. Therefore, $angle DCD'=angle DCB+angle BAD= 2pi-theta_A-theta_C-frac{pi}{3}=theta_B-frac{pi}{3}$. Therefore, angle opposite to $BD$ in $T = theta_B-frac{pi}{3}$.
Therefore, by symmetry, the angles of $T$ are $theta_A-frac{pi}{3}, theta_B-frac{pi}{3}$ and $theta_C-frac{pi}{3}$.
$blacksquare$
answered Jan 25 at 16:49
Anubhab GhosalAnubhab Ghosal
1,22319
$endgroup$
add a comment |
$begingroup$
Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.
Let $R$ denote a rotation by $frac{pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $triangle CDD'cong T$. As rotation preserves angles too, $angle BCD'=angle BR(A)R(D)=angle BAD$. Therefore, $angle DCD'=angle DCB+angle BAD= 2pi-theta_A-theta_C-frac{pi}{3}=theta_B-frac{pi}{3}$. Therefore, angle opposite to $BD$ in $T = theta_B-frac{pi}{3}$.
Therefore, by symmetry, the angles of $T$ are $theta_A-frac{pi}{3}, theta_B-frac{pi}{3}$ and $theta_C-frac{pi}{3}$.
$blacksquare$
answered Jan 25 at 16:49
Anubhab GhosalAnubhab Ghosal
1,22319
$endgroup$
add a comment |
$begingroup$
Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.
Let $R$ denote a rotation by $frac{pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $triangle CDD'cong T$. As rotation preserves angles too, $angle BCD'=angle BR(A)R(D)=angle BAD$. Therefore, $angle DCD'=angle DCB+angle BAD= 2pi-theta_A-theta_C-frac{pi}{3}=theta_B-frac{pi}{3}$. Therefore, angle opposite to $BD$ in $T = theta_B-frac{pi}{3}$.
Therefore, by symmetry, the angles of $T$ are $theta_A-frac{pi}{3}, theta_B-frac{pi}{3}$ and $theta_C-frac{pi}{3}$.
$blacksquare$
answered Jan 25 at 16:49
Anubhab GhosalAnubhab Ghosal
1,22319
$endgroup$
Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.
Let $R$ denote a rotation by $frac{pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $triangle CDD'cong T$. As rotation preserves angles too, $angle BCD'=angle BR(A)R(D)=angle BAD$. Therefore, $angle DCD'=angle DCB+angle BAD= 2pi-theta_A-theta_C-frac{pi}{3}=theta_B-frac{pi}{3}$. Therefore, angle opposite to $BD$ in $T = theta_B-frac{pi}{3}$.
Therefore, by symmetry, the angles of $T$ are $theta_A-frac{pi}{3}, theta_B-frac{pi}{3}$ and $theta_C-frac{pi}{3}$.
$blacksquare$
answered Jan 25 at 16:49
Anubhab GhosalAnubhab Ghosal
1,22319
answered Jan 25 at 16:49
Anubhab GhosalAnubhab Ghosal
1,22319
answered Jan 25 at 16:49
Anubhab GhosalAnubhab Ghosal
1,22319
answered Jan 25 at 16:49
Anubhab GhosalAnubhab Ghosal
1,22319
1,22319
add a comment |
add a comment |
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1
$begingroup$
Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
$endgroup$
– Ross Millikan
Jan 25 at 15:57
$begingroup$
But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
$endgroup$
– Umesh shankar
Jan 25 at 16:02