How do we solve this triangle using law of sines?
$begingroup$
ABC is a right triangle, $AB perp AC, DE perp BC, |AD| = |BD| = 3$. The area of ABC is $6$ times of the area of BDE. Evaluate $|AC|$.
How do we solve this triangle using law of sines?
geometry trigonometry
edited Dec 8 '18 at 14:32
Dylan
13.9k31127
asked Dec 8 '18 at 11:21
HamiltonHamilton
1778
$endgroup$
add a comment |
$begingroup$
ABC is a right triangle, $AB perp AC, DE perp BC, |AD| = |BD| = 3$. The area of ABC is $6$ times of the area of BDE. Evaluate $|AC|$.
How do we solve this triangle using law of sines?
geometry trigonometry
edited Dec 8 '18 at 14:32
Dylan
13.9k31127
asked Dec 8 '18 at 11:21
HamiltonHamilton
1778
$endgroup$
$begingroup$
There's easier ways than the sine rule. is that required?
$endgroup$
– TurlocTheRed
Dec 9 '18 at 4:05
add a comment |
$begingroup$
ABC is a right triangle, $AB perp AC, DE perp BC, |AD| = |BD| = 3$. The area of ABC is $6$ times of the area of BDE. Evaluate $|AC|$.
How do we solve this triangle using law of sines?
geometry trigonometry
edited Dec 8 '18 at 14:32
Dylan
13.9k31127
asked Dec 8 '18 at 11:21
HamiltonHamilton
1778
$endgroup$
ABC is a right triangle, $AB perp AC, DE perp BC, |AD| = |BD| = 3$. The area of ABC is $6$ times of the area of BDE. Evaluate $|AC|$.
How do we solve this triangle using law of sines?
geometry trigonometry
geometry trigonometry
edited Dec 8 '18 at 14:32
Dylan
13.9k31127
asked Dec 8 '18 at 11:21
HamiltonHamilton
1778
edited Dec 8 '18 at 14:32
Dylan
13.9k31127
asked Dec 8 '18 at 11:21
HamiltonHamilton
1778
edited Dec 8 '18 at 14:32
Dylan
13.9k31127
edited Dec 8 '18 at 14:32
Dylan
13.9k31127
edited Dec 8 '18 at 14:32
Dylan
13.9k31127
13.9k31127
asked Dec 8 '18 at 11:21
HamiltonHamilton
1778
asked Dec 8 '18 at 11:21
HamiltonHamilton
1778
asked Dec 8 '18 at 11:21
HamiltonHamilton
1778
1778
$begingroup$
There's easier ways than the sine rule. is that required?
$endgroup$
– TurlocTheRed
Dec 9 '18 at 4:05
add a comment |
$begingroup$
There's easier ways than the sine rule. is that required?
$endgroup$
– TurlocTheRed
Dec 9 '18 at 4:05
$begingroup$
There's easier ways than the sine rule. is that required?
$endgroup$
– TurlocTheRed
Dec 9 '18 at 4:05
$begingroup$
There's easier ways than the sine rule. is that required?
$endgroup$
– TurlocTheRed
Dec 9 '18 at 4:05
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let $angle ABC = theta$, then $EB = 3costheta$ and $AB=6$. Since the two triangles are similar, the area scale factor is
$$ frac{A_{BDE}}{A_{ABC}} = left(frac{EB}{AB}right)^2 = left(frac{costheta}{2}right)^2 = frac16 $$
$$ implies cos^2theta = frac23, sin^2theta = frac13, tan^2theta = frac12 $$
Finally
$$ AC = ABtantheta = frac{6}{sqrt{2}} = 3sqrt{2} $$
answered Dec 8 '18 at 11:33
DylanDylan
13.9k31127
$endgroup$
add a comment |
$begingroup$
We have $$A_{ABC}=6A_{BDE}$$ so we get
$$6frac{AC}{2}=6frac{DEcdot BE}{2}$$ with
$$frac{DE}{3}=frac{AC}{BC}$$ and $$frac{BE}{3}=frac{6}{BC}$$ we get
$$3AC=3cdot 3frac{AC}{BC}cdot frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$
answered Dec 8 '18 at 11:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
$begingroup$
Did you use sine law?
$endgroup$
– Hamilton
Dec 8 '18 at 13:46
$begingroup$
No i used the similarity of triangles.
$endgroup$
– Dr. Sonnhard Graubner
Dec 8 '18 at 14:36
add a comment |
$begingroup$
$triangle ABC$ and $triangle DEB$ are similar.$mangle ACB=mangle EDB$. As per sine law,$frac{c}{sinC}=frac{a}{sinA}=frac{b}{sinB}$ in $triangle ABC$. Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{e}{sinE}$in $triangle DEB$.
For the sake of convenience, $overline{AB}=c, overline{AC}=b,overline{BC}=a$ in $triangle ABC$ and $overline{DB}=e,overline{BE}=d,overline{DE}=b_1$ in $triangle DEB$.
$mangle E$ and $mangle A$ are $90^circ$ angle in $triangle DEB$ and $triangle ABC$ respectively.So, using sine law,
$frac{c}{sinC}=frac{b}{sinB}=frac{a}{sinA}=a$ because $sin{90^circ}=1$.
Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{3}{sinE}=3$ because $sin{E}=sin{90^circ}=1$
$sin{D}=sin{C}$.
So, ${d}=3sin{C}$ but $sin{C}=frac6a$.
So $d=frac{18}{a}$. But area of $triangle ABC$ is $6times$ the area of $triangle DEB$. It can be rewritten as $3b=3db_1$
So,$b=frac{18}{a}b_1$ $Rightarrow a*sin{B}=frac{54}{a}sin{B}$
$a^2=54 Rightarrow a=sqrt{54}=7.348469$
$a^2-c^2=b^2$ So,$54-36=18,Rightarrow b=sqrt{18}$
answered Dec 9 '18 at 3:43
Dhamnekar WinodDhamnekar Winod
427514
$endgroup$
add a comment |
$begingroup$
A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.
We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $sqrt{6}$:1.
We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.
By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself
answered Dec 9 '18 at 4:19
TurlocTheRedTurlocTheRed
916311
$endgroup$
$begingroup$
@TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 10:27
$begingroup$
More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
$endgroup$
– TurlocTheRed
Dec 9 '18 at 16:09
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $angle ABC = theta$, then $EB = 3costheta$ and $AB=6$. Since the two triangles are similar, the area scale factor is
$$ frac{A_{BDE}}{A_{ABC}} = left(frac{EB}{AB}right)^2 = left(frac{costheta}{2}right)^2 = frac16 $$
$$ implies cos^2theta = frac23, sin^2theta = frac13, tan^2theta = frac12 $$
Finally
$$ AC = ABtantheta = frac{6}{sqrt{2}} = 3sqrt{2} $$
answered Dec 8 '18 at 11:33
DylanDylan
13.9k31127
$endgroup$
add a comment |
$begingroup$
Let $angle ABC = theta$, then $EB = 3costheta$ and $AB=6$. Since the two triangles are similar, the area scale factor is
$$ frac{A_{BDE}}{A_{ABC}} = left(frac{EB}{AB}right)^2 = left(frac{costheta}{2}right)^2 = frac16 $$
$$ implies cos^2theta = frac23, sin^2theta = frac13, tan^2theta = frac12 $$
Finally
$$ AC = ABtantheta = frac{6}{sqrt{2}} = 3sqrt{2} $$
answered Dec 8 '18 at 11:33
DylanDylan
13.9k31127
$endgroup$
add a comment |
$begingroup$
Let $angle ABC = theta$, then $EB = 3costheta$ and $AB=6$. Since the two triangles are similar, the area scale factor is
$$ frac{A_{BDE}}{A_{ABC}} = left(frac{EB}{AB}right)^2 = left(frac{costheta}{2}right)^2 = frac16 $$
$$ implies cos^2theta = frac23, sin^2theta = frac13, tan^2theta = frac12 $$
Finally
$$ AC = ABtantheta = frac{6}{sqrt{2}} = 3sqrt{2} $$
answered Dec 8 '18 at 11:33
DylanDylan
13.9k31127
$endgroup$
Let $angle ABC = theta$, then $EB = 3costheta$ and $AB=6$. Since the two triangles are similar, the area scale factor is
$$ frac{A_{BDE}}{A_{ABC}} = left(frac{EB}{AB}right)^2 = left(frac{costheta}{2}right)^2 = frac16 $$
$$ implies cos^2theta = frac23, sin^2theta = frac13, tan^2theta = frac12 $$
Finally
$$ AC = ABtantheta = frac{6}{sqrt{2}} = 3sqrt{2} $$
answered Dec 8 '18 at 11:33
DylanDylan
13.9k31127
answered Dec 8 '18 at 11:33
DylanDylan
13.9k31127
answered Dec 8 '18 at 11:33
DylanDylan
13.9k31127
answered Dec 8 '18 at 11:33
DylanDylan
13.9k31127
13.9k31127
add a comment |
add a comment |
$begingroup$
We have $$A_{ABC}=6A_{BDE}$$ so we get
$$6frac{AC}{2}=6frac{DEcdot BE}{2}$$ with
$$frac{DE}{3}=frac{AC}{BC}$$ and $$frac{BE}{3}=frac{6}{BC}$$ we get
$$3AC=3cdot 3frac{AC}{BC}cdot frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$
answered Dec 8 '18 at 11:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
$begingroup$
Did you use sine law?
$endgroup$
– Hamilton
Dec 8 '18 at 13:46
$begingroup$
No i used the similarity of triangles.
$endgroup$
– Dr. Sonnhard Graubner
Dec 8 '18 at 14:36
add a comment |
$begingroup$
We have $$A_{ABC}=6A_{BDE}$$ so we get
$$6frac{AC}{2}=6frac{DEcdot BE}{2}$$ with
$$frac{DE}{3}=frac{AC}{BC}$$ and $$frac{BE}{3}=frac{6}{BC}$$ we get
$$3AC=3cdot 3frac{AC}{BC}cdot frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$
answered Dec 8 '18 at 11:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
$begingroup$
Did you use sine law?
$endgroup$
– Hamilton
Dec 8 '18 at 13:46
$begingroup$
No i used the similarity of triangles.
$endgroup$
– Dr. Sonnhard Graubner
Dec 8 '18 at 14:36
add a comment |
$begingroup$
We have $$A_{ABC}=6A_{BDE}$$ so we get
$$6frac{AC}{2}=6frac{DEcdot BE}{2}$$ with
$$frac{DE}{3}=frac{AC}{BC}$$ and $$frac{BE}{3}=frac{6}{BC}$$ we get
$$3AC=3cdot 3frac{AC}{BC}cdot frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$
answered Dec 8 '18 at 11:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
We have $$A_{ABC}=6A_{BDE}$$ so we get
$$6frac{AC}{2}=6frac{DEcdot BE}{2}$$ with
$$frac{DE}{3}=frac{AC}{BC}$$ and $$frac{BE}{3}=frac{6}{BC}$$ we get
$$3AC=3cdot 3frac{AC}{BC}cdot frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$
answered Dec 8 '18 at 11:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 8 '18 at 11:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 8 '18 at 11:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Dec 8 '18 at 11:34
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
77.8k42866
$begingroup$
Did you use sine law?
$endgroup$
– Hamilton
Dec 8 '18 at 13:46
$begingroup$
No i used the similarity of triangles.
$endgroup$
– Dr. Sonnhard Graubner
Dec 8 '18 at 14:36
add a comment |
$begingroup$
Did you use sine law?
$endgroup$
– Hamilton
Dec 8 '18 at 13:46
$begingroup$
No i used the similarity of triangles.
$endgroup$
– Dr. Sonnhard Graubner
Dec 8 '18 at 14:36
$begingroup$
Did you use sine law?
$endgroup$
– Hamilton
Dec 8 '18 at 13:46
$begingroup$
Did you use sine law?
$endgroup$
– Hamilton
Dec 8 '18 at 13:46
$begingroup$
No i used the similarity of triangles.
$endgroup$
– Dr. Sonnhard Graubner
Dec 8 '18 at 14:36
$begingroup$
No i used the similarity of triangles.
$endgroup$
– Dr. Sonnhard Graubner
Dec 8 '18 at 14:36
add a comment |
$begingroup$
$triangle ABC$ and $triangle DEB$ are similar.$mangle ACB=mangle EDB$. As per sine law,$frac{c}{sinC}=frac{a}{sinA}=frac{b}{sinB}$ in $triangle ABC$. Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{e}{sinE}$in $triangle DEB$.
For the sake of convenience, $overline{AB}=c, overline{AC}=b,overline{BC}=a$ in $triangle ABC$ and $overline{DB}=e,overline{BE}=d,overline{DE}=b_1$ in $triangle DEB$.
$mangle E$ and $mangle A$ are $90^circ$ angle in $triangle DEB$ and $triangle ABC$ respectively.So, using sine law,
$frac{c}{sinC}=frac{b}{sinB}=frac{a}{sinA}=a$ because $sin{90^circ}=1$.
Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{3}{sinE}=3$ because $sin{E}=sin{90^circ}=1$
$sin{D}=sin{C}$.
So, ${d}=3sin{C}$ but $sin{C}=frac6a$.
So $d=frac{18}{a}$. But area of $triangle ABC$ is $6times$ the area of $triangle DEB$. It can be rewritten as $3b=3db_1$
So,$b=frac{18}{a}b_1$ $Rightarrow a*sin{B}=frac{54}{a}sin{B}$
$a^2=54 Rightarrow a=sqrt{54}=7.348469$
$a^2-c^2=b^2$ So,$54-36=18,Rightarrow b=sqrt{18}$
answered Dec 9 '18 at 3:43
Dhamnekar WinodDhamnekar Winod
427514
$endgroup$
add a comment |
$begingroup$
$triangle ABC$ and $triangle DEB$ are similar.$mangle ACB=mangle EDB$. As per sine law,$frac{c}{sinC}=frac{a}{sinA}=frac{b}{sinB}$ in $triangle ABC$. Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{e}{sinE}$in $triangle DEB$.
For the sake of convenience, $overline{AB}=c, overline{AC}=b,overline{BC}=a$ in $triangle ABC$ and $overline{DB}=e,overline{BE}=d,overline{DE}=b_1$ in $triangle DEB$.
$mangle E$ and $mangle A$ are $90^circ$ angle in $triangle DEB$ and $triangle ABC$ respectively.So, using sine law,
$frac{c}{sinC}=frac{b}{sinB}=frac{a}{sinA}=a$ because $sin{90^circ}=1$.
Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{3}{sinE}=3$ because $sin{E}=sin{90^circ}=1$
$sin{D}=sin{C}$.
So, ${d}=3sin{C}$ but $sin{C}=frac6a$.
So $d=frac{18}{a}$. But area of $triangle ABC$ is $6times$ the area of $triangle DEB$. It can be rewritten as $3b=3db_1$
So,$b=frac{18}{a}b_1$ $Rightarrow a*sin{B}=frac{54}{a}sin{B}$
$a^2=54 Rightarrow a=sqrt{54}=7.348469$
$a^2-c^2=b^2$ So,$54-36=18,Rightarrow b=sqrt{18}$
answered Dec 9 '18 at 3:43
Dhamnekar WinodDhamnekar Winod
427514
$endgroup$
add a comment |
$begingroup$
$triangle ABC$ and $triangle DEB$ are similar.$mangle ACB=mangle EDB$. As per sine law,$frac{c}{sinC}=frac{a}{sinA}=frac{b}{sinB}$ in $triangle ABC$. Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{e}{sinE}$in $triangle DEB$.
For the sake of convenience, $overline{AB}=c, overline{AC}=b,overline{BC}=a$ in $triangle ABC$ and $overline{DB}=e,overline{BE}=d,overline{DE}=b_1$ in $triangle DEB$.
$mangle E$ and $mangle A$ are $90^circ$ angle in $triangle DEB$ and $triangle ABC$ respectively.So, using sine law,
$frac{c}{sinC}=frac{b}{sinB}=frac{a}{sinA}=a$ because $sin{90^circ}=1$.
Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{3}{sinE}=3$ because $sin{E}=sin{90^circ}=1$
$sin{D}=sin{C}$.
So, ${d}=3sin{C}$ but $sin{C}=frac6a$.
So $d=frac{18}{a}$. But area of $triangle ABC$ is $6times$ the area of $triangle DEB$. It can be rewritten as $3b=3db_1$
So,$b=frac{18}{a}b_1$ $Rightarrow a*sin{B}=frac{54}{a}sin{B}$
$a^2=54 Rightarrow a=sqrt{54}=7.348469$
$a^2-c^2=b^2$ So,$54-36=18,Rightarrow b=sqrt{18}$
answered Dec 9 '18 at 3:43
Dhamnekar WinodDhamnekar Winod
427514
$endgroup$
$triangle ABC$ and $triangle DEB$ are similar.$mangle ACB=mangle EDB$. As per sine law,$frac{c}{sinC}=frac{a}{sinA}=frac{b}{sinB}$ in $triangle ABC$. Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{e}{sinE}$in $triangle DEB$.
For the sake of convenience, $overline{AB}=c, overline{AC}=b,overline{BC}=a$ in $triangle ABC$ and $overline{DB}=e,overline{BE}=d,overline{DE}=b_1$ in $triangle DEB$.
$mangle E$ and $mangle A$ are $90^circ$ angle in $triangle DEB$ and $triangle ABC$ respectively.So, using sine law,
$frac{c}{sinC}=frac{b}{sinB}=frac{a}{sinA}=a$ because $sin{90^circ}=1$.
Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{3}{sinE}=3$ because $sin{E}=sin{90^circ}=1$
$sin{D}=sin{C}$.
So, ${d}=3sin{C}$ but $sin{C}=frac6a$.
So $d=frac{18}{a}$. But area of $triangle ABC$ is $6times$ the area of $triangle DEB$. It can be rewritten as $3b=3db_1$
So,$b=frac{18}{a}b_1$ $Rightarrow a*sin{B}=frac{54}{a}sin{B}$
$a^2=54 Rightarrow a=sqrt{54}=7.348469$
$a^2-c^2=b^2$ So,$54-36=18,Rightarrow b=sqrt{18}$
answered Dec 9 '18 at 3:43
Dhamnekar WinodDhamnekar Winod
427514
edited Dec 9 '18 at 3:53
answered Dec 9 '18 at 3:43
Dhamnekar WinodDhamnekar Winod
427514
answered Dec 9 '18 at 3:43
Dhamnekar WinodDhamnekar Winod
427514
answered Dec 9 '18 at 3:43
Dhamnekar WinodDhamnekar Winod
427514
427514
add a comment |
add a comment |
$begingroup$
A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.
We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $sqrt{6}$:1.
We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.
By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself
answered Dec 9 '18 at 4:19
TurlocTheRedTurlocTheRed
916311
$endgroup$
$begingroup$
@TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 10:27
$begingroup$
More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
$endgroup$
– TurlocTheRed
Dec 9 '18 at 16:09
add a comment |
$begingroup$
A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.
We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $sqrt{6}$:1.
We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.
By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself
answered Dec 9 '18 at 4:19
TurlocTheRedTurlocTheRed
916311
$endgroup$
$begingroup$
@TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 10:27
$begingroup$
More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
$endgroup$
– TurlocTheRed
Dec 9 '18 at 16:09
add a comment |
$begingroup$
A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.
We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $sqrt{6}$:1.
We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.
By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself
answered Dec 9 '18 at 4:19
TurlocTheRedTurlocTheRed
916311
$endgroup$
A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.
We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $sqrt{6}$:1.
We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.
By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself
answered Dec 9 '18 at 4:19
TurlocTheRedTurlocTheRed
916311
answered Dec 9 '18 at 4:19
TurlocTheRedTurlocTheRed
916311
answered Dec 9 '18 at 4:19
TurlocTheRedTurlocTheRed
916311
answered Dec 9 '18 at 4:19
TurlocTheRedTurlocTheRed
916311
916311
$begingroup$
@TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
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– Dhamnekar Winod
Dec 9 '18 at 10:27
$begingroup$
More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
$endgroup$
– TurlocTheRed
Dec 9 '18 at 16:09
add a comment |
$begingroup$
@TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 10:27
$begingroup$
More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
$endgroup$
– TurlocTheRed
Dec 9 '18 at 16:09
$begingroup$
@TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 10:27
$begingroup$
@TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
$endgroup$
– Dhamnekar Winod
Dec 9 '18 at 10:27
$begingroup$
More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
$endgroup$
– TurlocTheRed
Dec 9 '18 at 16:09
$begingroup$
More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
$endgroup$
– TurlocTheRed
Dec 9 '18 at 16:09
add a comment |
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There's easier ways than the sine rule. is that required?
$endgroup$
– TurlocTheRed
Dec 9 '18 at 4:05