Could you help me approach this problem using sine law?
$begingroup$
Given that ABC is an isosceles triangle, $[BD]$ is angle bisector, $angle BDA = 120^circ$. Evaluate the degree of $angle A $
Could you help me approach this problem using sine law?
Here's my attempt:
From the angle bisector theorem, we know that
$$dfrac{|AB|}{|BC|} = dfrac{|AD|}{|DC|} $$
In $triangle ADB$, let's call same angles $alpha$ and we have that
$$dfrac{|AB|}{sin (120)} = dfrac{|AD|}{sin(alpha )} implies dfrac{|AB|}{|AD|} = dfrac{sin (120)}{sin (alpha )} $$
This also equals
$$dfrac{|AB|}{|AD|} = dfrac{sin (120)}{sin (alpha )} = dfrac{|BC|}{|DC|}$$
Now $angle A 180-120-alpha = 60-alpha $, then
$$dfrac{|DB|}{sin (x)} = dfrac{|AB|}{sin (120)} implies dfrac{|DB|}{|AB|} = dfrac{sin(60-alpha )}{sin (120)} $$
triangle
asked Nov 30 '18 at 22:40
HamiltonHamilton
1798
$endgroup$
add a comment |
$begingroup$
Given that ABC is an isosceles triangle, $[BD]$ is angle bisector, $angle BDA = 120^circ$. Evaluate the degree of $angle A $
Could you help me approach this problem using sine law?
Here's my attempt:
From the angle bisector theorem, we know that
$$dfrac{|AB|}{|BC|} = dfrac{|AD|}{|DC|} $$
In $triangle ADB$, let's call same angles $alpha$ and we have that
$$dfrac{|AB|}{sin (120)} = dfrac{|AD|}{sin(alpha )} implies dfrac{|AB|}{|AD|} = dfrac{sin (120)}{sin (alpha )} $$
This also equals
$$dfrac{|AB|}{|AD|} = dfrac{sin (120)}{sin (alpha )} = dfrac{|BC|}{|DC|}$$
Now $angle A 180-120-alpha = 60-alpha $, then
$$dfrac{|DB|}{sin (x)} = dfrac{|AB|}{sin (120)} implies dfrac{|DB|}{|AB|} = dfrac{sin(60-alpha )}{sin (120)} $$
triangle
asked Nov 30 '18 at 22:40
HamiltonHamilton
1798
$endgroup$
1
$begingroup$
Evauate the degree of? angle in A I suppose
$endgroup$
– gimusi
Nov 30 '18 at 22:48
1
$begingroup$
@gimusi Yes, that's right. However, I want to use sine law.
$endgroup$
– Hamilton
Nov 30 '18 at 22:53
1
$begingroup$
ABC is isosceles, but you do not tell us which of the two sides are congruent. Can we assume $AB$ and $AC$ are congruent?
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:00
add a comment |
$begingroup$
Given that ABC is an isosceles triangle, $[BD]$ is angle bisector, $angle BDA = 120^circ$. Evaluate the degree of $angle A $
Could you help me approach this problem using sine law?
Here's my attempt:
From the angle bisector theorem, we know that
$$dfrac{|AB|}{|BC|} = dfrac{|AD|}{|DC|} $$
In $triangle ADB$, let's call same angles $alpha$ and we have that
$$dfrac{|AB|}{sin (120)} = dfrac{|AD|}{sin(alpha )} implies dfrac{|AB|}{|AD|} = dfrac{sin (120)}{sin (alpha )} $$
This also equals
$$dfrac{|AB|}{|AD|} = dfrac{sin (120)}{sin (alpha )} = dfrac{|BC|}{|DC|}$$
Now $angle A 180-120-alpha = 60-alpha $, then
$$dfrac{|DB|}{sin (x)} = dfrac{|AB|}{sin (120)} implies dfrac{|DB|}{|AB|} = dfrac{sin(60-alpha )}{sin (120)} $$
triangle
asked Nov 30 '18 at 22:40
HamiltonHamilton
1798
$endgroup$
Given that ABC is an isosceles triangle, $[BD]$ is angle bisector, $angle BDA = 120^circ$. Evaluate the degree of $angle A $
Could you help me approach this problem using sine law?
Here's my attempt:
From the angle bisector theorem, we know that
$$dfrac{|AB|}{|BC|} = dfrac{|AD|}{|DC|} $$
In $triangle ADB$, let's call same angles $alpha$ and we have that
$$dfrac{|AB|}{sin (120)} = dfrac{|AD|}{sin(alpha )} implies dfrac{|AB|}{|AD|} = dfrac{sin (120)}{sin (alpha )} $$
This also equals
$$dfrac{|AB|}{|AD|} = dfrac{sin (120)}{sin (alpha )} = dfrac{|BC|}{|DC|}$$
Now $angle A 180-120-alpha = 60-alpha $, then
$$dfrac{|DB|}{sin (x)} = dfrac{|AB|}{sin (120)} implies dfrac{|DB|}{|AB|} = dfrac{sin(60-alpha )}{sin (120)} $$
triangle
triangle
asked Nov 30 '18 at 22:40
HamiltonHamilton
1798
asked Nov 30 '18 at 22:40
HamiltonHamilton
1798
edited Nov 30 '18 at 22:53
Hamilton
asked Nov 30 '18 at 22:40
HamiltonHamilton
1798
asked Nov 30 '18 at 22:40
HamiltonHamilton
1798
asked Nov 30 '18 at 22:40
HamiltonHamilton
1798
1798
1
$begingroup$
Evauate the degree of? angle in A I suppose
$endgroup$
– gimusi
Nov 30 '18 at 22:48
1
$begingroup$
@gimusi Yes, that's right. However, I want to use sine law.
$endgroup$
– Hamilton
Nov 30 '18 at 22:53
1
$begingroup$
ABC is isosceles, but you do not tell us which of the two sides are congruent. Can we assume $AB$ and $AC$ are congruent?
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:00
add a comment |
1
$begingroup$
Evauate the degree of? angle in A I suppose
$endgroup$
– gimusi
Nov 30 '18 at 22:48
1
$begingroup$
@gimusi Yes, that's right. However, I want to use sine law.
$endgroup$
– Hamilton
Nov 30 '18 at 22:53
1
$begingroup$
ABC is isosceles, but you do not tell us which of the two sides are congruent. Can we assume $AB$ and $AC$ are congruent?
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:00
1
1
$begingroup$
Evauate the degree of? angle in A I suppose
$endgroup$
– gimusi
Nov 30 '18 at 22:48
$begingroup$
Evauate the degree of? angle in A I suppose
$endgroup$
– gimusi
Nov 30 '18 at 22:48
1
1
$begingroup$
@gimusi Yes, that's right. However, I want to use sine law.
$endgroup$
– Hamilton
Nov 30 '18 at 22:53
$begingroup$
@gimusi Yes, that's right. However, I want to use sine law.
$endgroup$
– Hamilton
Nov 30 '18 at 22:53
1
1
$begingroup$
ABC is isosceles, but you do not tell us which of the two sides are congruent. Can we assume $AB$ and $AC$ are congruent?
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:00
$begingroup$
ABC is isosceles, but you do not tell us which of the two sides are congruent. Can we assume $AB$ and $AC$ are congruent?
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:00
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I think Law of Sines is ill-suited for this problem. However, in order to understand this, you first need to see the answer. First, $AB=AC$ since triangle $ABC$ is isosceles. Therefore:
$$angle ABC=angle ACB$$
Also, $BC$ is an angle bisector, so:
$$angle ABD=frac 1 2angle ABC$$
From triangle $ABD$, we have:
$$angle A+frac 1 2angle ABC+120^circ=180^circ$$
From triangle $ABC$, we have:
$$angle A+2angle ABC=180^circ$$
Subtract the second equation by the first:
$$frac 3 2angle ABC-120^circ=0rightarrow angle ABC=80^circ$$
Substitute back into the second equation:
$$angle A+160^circ=180^circrightarrowangle A=20^circ$$
Thus, our final answer is $20^circ$. This means:
$$sin A=frac i 2left(sqrt[3]{frac{-1-sqrt{-3}}{2}}-sqrt[3]{frac{-1+sqrt{-3}}{2}}right)$$
I think it would be very hard to derive this complex expression from using Law of Sines in order to solve for A, which is why Law of Sines is not a good way to solve this rather simple angle problem.
answered Nov 30 '18 at 23:25
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
Let $x$ be the measure of $angle ABD$
and $y$ be the measure of $angle BAC$
$x+y+ 120 = 180\
4x + y = 180$
And solve the system of equations.
Regrading law of sines.. that might be useful you knew more side lengths. Right now all you know is that the triangle is isosceles.
answered Nov 30 '18 at 23:39
Doug MDoug M
44.9k31854
$endgroup$
add a comment |
$begingroup$
We have that since $angle BDC=60° implies angle ACB=angle ABC=80*$ and $angle BAC=20°$.
To find the result by law of sine let
- $BC=a $
- $AB=AC=b$
- $BD=x$
- $DC=y$
- $angle BAC=alpha$
then we have to solve the following systems of $5$ equations in $5$ unknowns
- $frac{a}{sin alpha}=frac{b}{sin 80}$
- $frac{a}{sin 60}=frac{x}{sin 80}=frac{y}{sin 40}$
- $frac{x}{sin alpha}=frac{b}{sin 120}=frac{b-y}{sin 40}$
answered Nov 30 '18 at 23:05
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Where does the 40 come from?
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:10
$begingroup$
@NobleMushtak $angle BDC=60°$
$endgroup$
– gimusi
Nov 30 '18 at 23:11
$begingroup$
That does not really make it obvious how you got the $40^circ$ using Law of Sines.
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:26
$begingroup$
Also, the second equation should have $sin alpha$, not $sin A$. $alpha$ is the measure of $angle ABD$ and $angle BDC$.
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:27
$begingroup$
@NobleMushtak The triangle is isosceles, therefore $angle B=angle C$ and since $B/2+C+60=180 implies B=C=80$.
$endgroup$
– gimusi
Nov 30 '18 at 23:28
|
show 3 more comments
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think Law of Sines is ill-suited for this problem. However, in order to understand this, you first need to see the answer. First, $AB=AC$ since triangle $ABC$ is isosceles. Therefore:
$$angle ABC=angle ACB$$
Also, $BC$ is an angle bisector, so:
$$angle ABD=frac 1 2angle ABC$$
From triangle $ABD$, we have:
$$angle A+frac 1 2angle ABC+120^circ=180^circ$$
From triangle $ABC$, we have:
$$angle A+2angle ABC=180^circ$$
Subtract the second equation by the first:
$$frac 3 2angle ABC-120^circ=0rightarrow angle ABC=80^circ$$
Substitute back into the second equation:
$$angle A+160^circ=180^circrightarrowangle A=20^circ$$
Thus, our final answer is $20^circ$. This means:
$$sin A=frac i 2left(sqrt[3]{frac{-1-sqrt{-3}}{2}}-sqrt[3]{frac{-1+sqrt{-3}}{2}}right)$$
I think it would be very hard to derive this complex expression from using Law of Sines in order to solve for A, which is why Law of Sines is not a good way to solve this rather simple angle problem.
answered Nov 30 '18 at 23:25
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
I think Law of Sines is ill-suited for this problem. However, in order to understand this, you first need to see the answer. First, $AB=AC$ since triangle $ABC$ is isosceles. Therefore:
$$angle ABC=angle ACB$$
Also, $BC$ is an angle bisector, so:
$$angle ABD=frac 1 2angle ABC$$
From triangle $ABD$, we have:
$$angle A+frac 1 2angle ABC+120^circ=180^circ$$
From triangle $ABC$, we have:
$$angle A+2angle ABC=180^circ$$
Subtract the second equation by the first:
$$frac 3 2angle ABC-120^circ=0rightarrow angle ABC=80^circ$$
Substitute back into the second equation:
$$angle A+160^circ=180^circrightarrowangle A=20^circ$$
Thus, our final answer is $20^circ$. This means:
$$sin A=frac i 2left(sqrt[3]{frac{-1-sqrt{-3}}{2}}-sqrt[3]{frac{-1+sqrt{-3}}{2}}right)$$
I think it would be very hard to derive this complex expression from using Law of Sines in order to solve for A, which is why Law of Sines is not a good way to solve this rather simple angle problem.
answered Nov 30 '18 at 23:25
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
I think Law of Sines is ill-suited for this problem. However, in order to understand this, you first need to see the answer. First, $AB=AC$ since triangle $ABC$ is isosceles. Therefore:
$$angle ABC=angle ACB$$
Also, $BC$ is an angle bisector, so:
$$angle ABD=frac 1 2angle ABC$$
From triangle $ABD$, we have:
$$angle A+frac 1 2angle ABC+120^circ=180^circ$$
From triangle $ABC$, we have:
$$angle A+2angle ABC=180^circ$$
Subtract the second equation by the first:
$$frac 3 2angle ABC-120^circ=0rightarrow angle ABC=80^circ$$
Substitute back into the second equation:
$$angle A+160^circ=180^circrightarrowangle A=20^circ$$
Thus, our final answer is $20^circ$. This means:
$$sin A=frac i 2left(sqrt[3]{frac{-1-sqrt{-3}}{2}}-sqrt[3]{frac{-1+sqrt{-3}}{2}}right)$$
I think it would be very hard to derive this complex expression from using Law of Sines in order to solve for A, which is why Law of Sines is not a good way to solve this rather simple angle problem.
answered Nov 30 '18 at 23:25
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
I think Law of Sines is ill-suited for this problem. However, in order to understand this, you first need to see the answer. First, $AB=AC$ since triangle $ABC$ is isosceles. Therefore:
$$angle ABC=angle ACB$$
Also, $BC$ is an angle bisector, so:
$$angle ABD=frac 1 2angle ABC$$
From triangle $ABD$, we have:
$$angle A+frac 1 2angle ABC+120^circ=180^circ$$
From triangle $ABC$, we have:
$$angle A+2angle ABC=180^circ$$
Subtract the second equation by the first:
$$frac 3 2angle ABC-120^circ=0rightarrow angle ABC=80^circ$$
Substitute back into the second equation:
$$angle A+160^circ=180^circrightarrowangle A=20^circ$$
Thus, our final answer is $20^circ$. This means:
$$sin A=frac i 2left(sqrt[3]{frac{-1-sqrt{-3}}{2}}-sqrt[3]{frac{-1+sqrt{-3}}{2}}right)$$
I think it would be very hard to derive this complex expression from using Law of Sines in order to solve for A, which is why Law of Sines is not a good way to solve this rather simple angle problem.
answered Nov 30 '18 at 23:25
Noble MushtakNoble Mushtak
15.3k1835
answered Nov 30 '18 at 23:25
Noble MushtakNoble Mushtak
15.3k1835
answered Nov 30 '18 at 23:25
Noble MushtakNoble Mushtak
15.3k1835
answered Nov 30 '18 at 23:25
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
add a comment |
$begingroup$
Let $x$ be the measure of $angle ABD$
and $y$ be the measure of $angle BAC$
$x+y+ 120 = 180\
4x + y = 180$
And solve the system of equations.
Regrading law of sines.. that might be useful you knew more side lengths. Right now all you know is that the triangle is isosceles.
answered Nov 30 '18 at 23:39
Doug MDoug M
44.9k31854
$endgroup$
add a comment |
$begingroup$
Let $x$ be the measure of $angle ABD$
and $y$ be the measure of $angle BAC$
$x+y+ 120 = 180\
4x + y = 180$
And solve the system of equations.
Regrading law of sines.. that might be useful you knew more side lengths. Right now all you know is that the triangle is isosceles.
answered Nov 30 '18 at 23:39
Doug MDoug M
44.9k31854
$endgroup$
add a comment |
$begingroup$
Let $x$ be the measure of $angle ABD$
and $y$ be the measure of $angle BAC$
$x+y+ 120 = 180\
4x + y = 180$
And solve the system of equations.
Regrading law of sines.. that might be useful you knew more side lengths. Right now all you know is that the triangle is isosceles.
answered Nov 30 '18 at 23:39
Doug MDoug M
44.9k31854
$endgroup$
Let $x$ be the measure of $angle ABD$
and $y$ be the measure of $angle BAC$
$x+y+ 120 = 180\
4x + y = 180$
And solve the system of equations.
Regrading law of sines.. that might be useful you knew more side lengths. Right now all you know is that the triangle is isosceles.
answered Nov 30 '18 at 23:39
Doug MDoug M
44.9k31854
answered Nov 30 '18 at 23:39
Doug MDoug M
44.9k31854
answered Nov 30 '18 at 23:39
Doug MDoug M
44.9k31854
answered Nov 30 '18 at 23:39
Doug MDoug M
44.9k31854
44.9k31854
add a comment |
add a comment |
$begingroup$
We have that since $angle BDC=60° implies angle ACB=angle ABC=80*$ and $angle BAC=20°$.
To find the result by law of sine let
- $BC=a $
- $AB=AC=b$
- $BD=x$
- $DC=y$
- $angle BAC=alpha$
then we have to solve the following systems of $5$ equations in $5$ unknowns
- $frac{a}{sin alpha}=frac{b}{sin 80}$
- $frac{a}{sin 60}=frac{x}{sin 80}=frac{y}{sin 40}$
- $frac{x}{sin alpha}=frac{b}{sin 120}=frac{b-y}{sin 40}$
answered Nov 30 '18 at 23:05
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Where does the 40 come from?
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:10
$begingroup$
@NobleMushtak $angle BDC=60°$
$endgroup$
– gimusi
Nov 30 '18 at 23:11
$begingroup$
That does not really make it obvious how you got the $40^circ$ using Law of Sines.
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:26
$begingroup$
Also, the second equation should have $sin alpha$, not $sin A$. $alpha$ is the measure of $angle ABD$ and $angle BDC$.
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:27
$begingroup$
@NobleMushtak The triangle is isosceles, therefore $angle B=angle C$ and since $B/2+C+60=180 implies B=C=80$.
$endgroup$
– gimusi
Nov 30 '18 at 23:28
|
show 3 more comments
$begingroup$
We have that since $angle BDC=60° implies angle ACB=angle ABC=80*$ and $angle BAC=20°$.
To find the result by law of sine let
- $BC=a $
- $AB=AC=b$
- $BD=x$
- $DC=y$
- $angle BAC=alpha$
then we have to solve the following systems of $5$ equations in $5$ unknowns
- $frac{a}{sin alpha}=frac{b}{sin 80}$
- $frac{a}{sin 60}=frac{x}{sin 80}=frac{y}{sin 40}$
- $frac{x}{sin alpha}=frac{b}{sin 120}=frac{b-y}{sin 40}$
answered Nov 30 '18 at 23:05
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Where does the 40 come from?
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:10
$begingroup$
@NobleMushtak $angle BDC=60°$
$endgroup$
– gimusi
Nov 30 '18 at 23:11
$begingroup$
That does not really make it obvious how you got the $40^circ$ using Law of Sines.
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:26
$begingroup$
Also, the second equation should have $sin alpha$, not $sin A$. $alpha$ is the measure of $angle ABD$ and $angle BDC$.
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:27
$begingroup$
@NobleMushtak The triangle is isosceles, therefore $angle B=angle C$ and since $B/2+C+60=180 implies B=C=80$.
$endgroup$
– gimusi
Nov 30 '18 at 23:28
|
show 3 more comments
$begingroup$
We have that since $angle BDC=60° implies angle ACB=angle ABC=80*$ and $angle BAC=20°$.
To find the result by law of sine let
- $BC=a $
- $AB=AC=b$
- $BD=x$
- $DC=y$
- $angle BAC=alpha$
then we have to solve the following systems of $5$ equations in $5$ unknowns
- $frac{a}{sin alpha}=frac{b}{sin 80}$
- $frac{a}{sin 60}=frac{x}{sin 80}=frac{y}{sin 40}$
- $frac{x}{sin alpha}=frac{b}{sin 120}=frac{b-y}{sin 40}$
answered Nov 30 '18 at 23:05
gimusigimusi
92.8k84494
$endgroup$
We have that since $angle BDC=60° implies angle ACB=angle ABC=80*$ and $angle BAC=20°$.
To find the result by law of sine let
- $BC=a $
- $AB=AC=b$
- $BD=x$
- $DC=y$
- $angle BAC=alpha$
then we have to solve the following systems of $5$ equations in $5$ unknowns
- $frac{a}{sin alpha}=frac{b}{sin 80}$
- $frac{a}{sin 60}=frac{x}{sin 80}=frac{y}{sin 40}$
- $frac{x}{sin alpha}=frac{b}{sin 120}=frac{b-y}{sin 40}$
answered Nov 30 '18 at 23:05
gimusigimusi
92.8k84494
edited Dec 1 '18 at 8:23
answered Nov 30 '18 at 23:05
gimusigimusi
92.8k84494
answered Nov 30 '18 at 23:05
gimusigimusi
92.8k84494
answered Nov 30 '18 at 23:05
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
Where does the 40 come from?
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:10
$begingroup$
@NobleMushtak $angle BDC=60°$
$endgroup$
– gimusi
Nov 30 '18 at 23:11
$begingroup$
That does not really make it obvious how you got the $40^circ$ using Law of Sines.
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:26
$begingroup$
Also, the second equation should have $sin alpha$, not $sin A$. $alpha$ is the measure of $angle ABD$ and $angle BDC$.
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:27
$begingroup$
@NobleMushtak The triangle is isosceles, therefore $angle B=angle C$ and since $B/2+C+60=180 implies B=C=80$.
$endgroup$
– gimusi
Nov 30 '18 at 23:28
|
show 3 more comments
$begingroup$
Where does the 40 come from?
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:10
$begingroup$
@NobleMushtak $angle BDC=60°$
$endgroup$
– gimusi
Nov 30 '18 at 23:11
$begingroup$
That does not really make it obvious how you got the $40^circ$ using Law of Sines.
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:26
$begingroup$
Also, the second equation should have $sin alpha$, not $sin A$. $alpha$ is the measure of $angle ABD$ and $angle BDC$.
$endgroup$
– Noble Mushtak
Nov 30 '18 at 23:27
$begingroup$
@NobleMushtak The triangle is isosceles, therefore $angle B=angle C$ and since $B/2+C+60=180 implies B=C=80$.
$endgroup$
– gimusi
Nov 30 '18 at 23:28
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Where does the 40 come from?
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– Noble Mushtak
Nov 30 '18 at 23:10
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Where does the 40 come from?
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– Noble Mushtak
Nov 30 '18 at 23:10
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@NobleMushtak $angle BDC=60°$
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– gimusi
Nov 30 '18 at 23:11
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@NobleMushtak $angle BDC=60°$
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– gimusi
Nov 30 '18 at 23:11
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That does not really make it obvious how you got the $40^circ$ using Law of Sines.
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– Noble Mushtak
Nov 30 '18 at 23:26
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That does not really make it obvious how you got the $40^circ$ using Law of Sines.
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– Noble Mushtak
Nov 30 '18 at 23:26
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Also, the second equation should have $sin alpha$, not $sin A$. $alpha$ is the measure of $angle ABD$ and $angle BDC$.
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– Noble Mushtak
Nov 30 '18 at 23:27
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Also, the second equation should have $sin alpha$, not $sin A$. $alpha$ is the measure of $angle ABD$ and $angle BDC$.
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– Noble Mushtak
Nov 30 '18 at 23:27
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@NobleMushtak The triangle is isosceles, therefore $angle B=angle C$ and since $B/2+C+60=180 implies B=C=80$.
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– gimusi
Nov 30 '18 at 23:28
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@NobleMushtak The triangle is isosceles, therefore $angle B=angle C$ and since $B/2+C+60=180 implies B=C=80$.
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– gimusi
Nov 30 '18 at 23:28
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Evauate the degree of? angle in A I suppose
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– gimusi
Nov 30 '18 at 22:48
1
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@gimusi Yes, that's right. However, I want to use sine law.
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– Hamilton
Nov 30 '18 at 22:53
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ABC is isosceles, but you do not tell us which of the two sides are congruent. Can we assume $AB$ and $AC$ are congruent?
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– Noble Mushtak
Nov 30 '18 at 23:00