Mixing
Finding precise measurement of angles using algebra or arithmetic.
$begingroup$
SchematicFor a $triangle ABC$ there exist three sides, $a, b, c$ and by using the formulas of the altitudes,I am able to find all 3 angles with the help of trigonometric identities SohCahToa.
At the same time, by using three angles I'm able to find the lenghts of all triangle without the help of the Law of sine.
The following equations represent a schematic of all triangles to obtain the result needed to find 3 sides and three angles of triangles.We can use the Law of Cosine for all three to get the three angles also.
$(ctimessin B)^2+( btimescos C)^2=b^2$
$(ctimessin A)^2+(atimescos C)^2=a^2$
$(ctimessin B)^2+(ctimescos B)^2=c^2$
$(ctimessin A)^2+(ctimescos A)^2=c^2$
$(btimescos C)+(ctimescos B)=a$
$(atimescos C)+(ctimescos A)=b$
$(atimescos B)+(b timescos A)=c$
$(ctimescos B)+(b timescos C)=a$
$(ctimescos A)+(atimescos C)=b$
and the formula to find the altitudes for all triangles:
$h_a=frac{sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$
$h_b=frac{sqrt{(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}$
$h_c=frac{sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}$
I have three altitudes and all three sides and I'm able to find all three angles with the help of SohCahToa without the Law of cosine even though I have three sides without three angles.
$(sin A)^2+(sin B)^2+(sin C)^2+(cos A)^2+(cos B)^2+(cos C)^2=3$ they apply for all triangles.
And the law of sine states when one of the base of the triangle is one then:
$frac{sin A}{a}=frac{sin B}{b}=frac{sin C}{c}=frac{1}{d}=$ where the reciprocal of the diameter is $angle C$
$frac{a}{sin A}=frac{b}{sin B}=frac{c}{sin C}=d$ where d is the diameter
Let $triangle ABC$ be sides $a$, $b$, and $c$. I am using that the length of the side opposite vertex $A$ is called $a$, $B$ for $b$ and $C$ for $c$.
If $theta=angle C$. Then the Cosine says that
$c^2=a^2+b^2-2abcos theta$.
Since we know $a$, $b$, and $c$, we can use the formula for all three altitudes or above formula to calculate $costheta$.
We can use the Law of Cosine and the formula of three altitudes to get all six angles.
geometry trigonometry proof-verification logic
asked Jan 31 at 11:25
user136391user136391
76
$endgroup$
add a comment |
$begingroup$
SchematicFor a $triangle ABC$ there exist three sides, $a, b, c$ and by using the formulas of the altitudes,I am able to find all 3 angles with the help of trigonometric identities SohCahToa.
At the same time, by using three angles I'm able to find the lenghts of all triangle without the help of the Law of sine.
The following equations represent a schematic of all triangles to obtain the result needed to find 3 sides and three angles of triangles.We can use the Law of Cosine for all three to get the three angles also.
$(ctimessin B)^2+( btimescos C)^2=b^2$
$(ctimessin A)^2+(atimescos C)^2=a^2$
$(ctimessin B)^2+(ctimescos B)^2=c^2$
$(ctimessin A)^2+(ctimescos A)^2=c^2$
$(btimescos C)+(ctimescos B)=a$
$(atimescos C)+(ctimescos A)=b$
$(atimescos B)+(b timescos A)=c$
$(ctimescos B)+(b timescos C)=a$
$(ctimescos A)+(atimescos C)=b$
and the formula to find the altitudes for all triangles:
$h_a=frac{sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$
$h_b=frac{sqrt{(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}$
$h_c=frac{sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}$
I have three altitudes and all three sides and I'm able to find all three angles with the help of SohCahToa without the Law of cosine even though I have three sides without three angles.
$(sin A)^2+(sin B)^2+(sin C)^2+(cos A)^2+(cos B)^2+(cos C)^2=3$ they apply for all triangles.
And the law of sine states when one of the base of the triangle is one then:
$frac{sin A}{a}=frac{sin B}{b}=frac{sin C}{c}=frac{1}{d}=$ where the reciprocal of the diameter is $angle C$
$frac{a}{sin A}=frac{b}{sin B}=frac{c}{sin C}=d$ where d is the diameter
Let $triangle ABC$ be sides $a$, $b$, and $c$. I am using that the length of the side opposite vertex $A$ is called $a$, $B$ for $b$ and $C$ for $c$.
If $theta=angle C$. Then the Cosine says that
$c^2=a^2+b^2-2abcos theta$.
Since we know $a$, $b$, and $c$, we can use the formula for all three altitudes or above formula to calculate $costheta$.
We can use the Law of Cosine and the formula of three altitudes to get all six angles.
geometry trigonometry proof-verification logic
asked Jan 31 at 11:25
user136391user136391
76
$endgroup$
$begingroup$
What is your answer?
$endgroup$
– Michael Rozenberg
Jan 31 at 11:39
$begingroup$
@Michael Rozenberg I've added a simple picture guide and I'm alone to interpret whether I am right.
$endgroup$
– user136391
Jan 31 at 11:50
$begingroup$
I think, you are right.
$endgroup$
– Michael Rozenberg
Jan 31 at 11:52
add a comment |
$begingroup$
SchematicFor a $triangle ABC$ there exist three sides, $a, b, c$ and by using the formulas of the altitudes,I am able to find all 3 angles with the help of trigonometric identities SohCahToa.
At the same time, by using three angles I'm able to find the lenghts of all triangle without the help of the Law of sine.
The following equations represent a schematic of all triangles to obtain the result needed to find 3 sides and three angles of triangles.We can use the Law of Cosine for all three to get the three angles also.
$(ctimessin B)^2+( btimescos C)^2=b^2$
$(ctimessin A)^2+(atimescos C)^2=a^2$
$(ctimessin B)^2+(ctimescos B)^2=c^2$
$(ctimessin A)^2+(ctimescos A)^2=c^2$
$(btimescos C)+(ctimescos B)=a$
$(atimescos C)+(ctimescos A)=b$
$(atimescos B)+(b timescos A)=c$
$(ctimescos B)+(b timescos C)=a$
$(ctimescos A)+(atimescos C)=b$
and the formula to find the altitudes for all triangles:
$h_a=frac{sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$
$h_b=frac{sqrt{(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}$
$h_c=frac{sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}$
I have three altitudes and all three sides and I'm able to find all three angles with the help of SohCahToa without the Law of cosine even though I have three sides without three angles.
$(sin A)^2+(sin B)^2+(sin C)^2+(cos A)^2+(cos B)^2+(cos C)^2=3$ they apply for all triangles.
And the law of sine states when one of the base of the triangle is one then:
$frac{sin A}{a}=frac{sin B}{b}=frac{sin C}{c}=frac{1}{d}=$ where the reciprocal of the diameter is $angle C$
$frac{a}{sin A}=frac{b}{sin B}=frac{c}{sin C}=d$ where d is the diameter
Let $triangle ABC$ be sides $a$, $b$, and $c$. I am using that the length of the side opposite vertex $A$ is called $a$, $B$ for $b$ and $C$ for $c$.
If $theta=angle C$. Then the Cosine says that
$c^2=a^2+b^2-2abcos theta$.
Since we know $a$, $b$, and $c$, we can use the formula for all three altitudes or above formula to calculate $costheta$.
We can use the Law of Cosine and the formula of three altitudes to get all six angles.
geometry trigonometry proof-verification logic
asked Jan 31 at 11:25
user136391user136391
76
$endgroup$
SchematicFor a $triangle ABC$ there exist three sides, $a, b, c$ and by using the formulas of the altitudes,I am able to find all 3 angles with the help of trigonometric identities SohCahToa.
At the same time, by using three angles I'm able to find the lenghts of all triangle without the help of the Law of sine.
The following equations represent a schematic of all triangles to obtain the result needed to find 3 sides and three angles of triangles.We can use the Law of Cosine for all three to get the three angles also.
$(ctimessin B)^2+( btimescos C)^2=b^2$
$(ctimessin A)^2+(atimescos C)^2=a^2$
$(ctimessin B)^2+(ctimescos B)^2=c^2$
$(ctimessin A)^2+(ctimescos A)^2=c^2$
$(btimescos C)+(ctimescos B)=a$
$(atimescos C)+(ctimescos A)=b$
$(atimescos B)+(b timescos A)=c$
$(ctimescos B)+(b timescos C)=a$
$(ctimescos A)+(atimescos C)=b$
and the formula to find the altitudes for all triangles:
$h_a=frac{sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$
$h_b=frac{sqrt{(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2b}$
$h_c=frac{sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2c}$
I have three altitudes and all three sides and I'm able to find all three angles with the help of SohCahToa without the Law of cosine even though I have three sides without three angles.
$(sin A)^2+(sin B)^2+(sin C)^2+(cos A)^2+(cos B)^2+(cos C)^2=3$ they apply for all triangles.
And the law of sine states when one of the base of the triangle is one then:
$frac{sin A}{a}=frac{sin B}{b}=frac{sin C}{c}=frac{1}{d}=$ where the reciprocal of the diameter is $angle C$
$frac{a}{sin A}=frac{b}{sin B}=frac{c}{sin C}=d$ where d is the diameter
Let $triangle ABC$ be sides $a$, $b$, and $c$. I am using that the length of the side opposite vertex $A$ is called $a$, $B$ for $b$ and $C$ for $c$.
If $theta=angle C$. Then the Cosine says that
$c^2=a^2+b^2-2abcos theta$.
Since we know $a$, $b$, and $c$, we can use the formula for all three altitudes or above formula to calculate $costheta$.
We can use the Law of Cosine and the formula of three altitudes to get all six angles.
geometry trigonometry proof-verification logic
geometry trigonometry proof-verification logic
asked Jan 31 at 11:25
user136391user136391
76
asked Jan 31 at 11:25
user136391user136391
76
edited Feb 4 at 9:17
user136391
asked Jan 31 at 11:25
user136391user136391
76
asked Jan 31 at 11:25
user136391user136391
76
asked Jan 31 at 11:25
user136391user136391
76
76
$begingroup$
What is your answer?
$endgroup$
– Michael Rozenberg
Jan 31 at 11:39
$begingroup$
@Michael Rozenberg I've added a simple picture guide and I'm alone to interpret whether I am right.
$endgroup$
– user136391
Jan 31 at 11:50
$begingroup$
I think, you are right.
$endgroup$
– Michael Rozenberg
Jan 31 at 11:52
add a comment |
$begingroup$
What is your answer?
$endgroup$
– Michael Rozenberg
Jan 31 at 11:39
$begingroup$
@Michael Rozenberg I've added a simple picture guide and I'm alone to interpret whether I am right.
$endgroup$
– user136391
Jan 31 at 11:50
$begingroup$
I think, you are right.
$endgroup$
– Michael Rozenberg
Jan 31 at 11:52
$begingroup$
What is your answer?
$endgroup$
– Michael Rozenberg
Jan 31 at 11:39
$begingroup$
What is your answer?
$endgroup$
– Michael Rozenberg
Jan 31 at 11:39
$begingroup$
@Michael Rozenberg I've added a simple picture guide and I'm alone to interpret whether I am right.
$endgroup$
– user136391
Jan 31 at 11:50
$begingroup$
@Michael Rozenberg I've added a simple picture guide and I'm alone to interpret whether I am right.
$endgroup$
– user136391
Jan 31 at 11:50
$begingroup$
I think, you are right.
$endgroup$
– Michael Rozenberg
Jan 31 at 11:52
$begingroup$
I think, you are right.
$endgroup$
– Michael Rozenberg
Jan 31 at 11:52
add a comment |
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$begingroup$
What is your answer?
$endgroup$
– Michael Rozenberg
Jan 31 at 11:39
$begingroup$
@Michael Rozenberg I've added a simple picture guide and I'm alone to interpret whether I am right.
$endgroup$
– user136391
Jan 31 at 11:50
$begingroup$
I think, you are right.
$endgroup$
– Michael Rozenberg
Jan 31 at 11:52