Mixing
In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, $D$ is on $AC$, with $BC=AD$. Find $angle DBC$....
$begingroup$
In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, point $D$ is on $AC$, with $BC=AD$. Find $angle DBC$.
I know the correct solution, but I'm more interested in where is the problem in my solution.
My solution :
Now in $triangle ABD$, applying the sine rule:
$$frac{AD}{sinalpha} = frac{BD}{sin 20^circ} tag{1}$$
In $triangle BDC$:
$$frac{BD}{sin 80^circ} = frac{BC}{sin(180^circ-beta)} tag{2}$$
We know $AD= BC$; put in to $(1)$:
$$frac{BC}{sinalpha} = frac{BD}{sin 20^circ} tag{3}$$
Comparing $(2)$ and $(3)$:
$$frac{BC}{BD} = frac{sinalpha}{sin 20^circ} = frac{sin(180^circ-beta)}{sin 80^circ} tag{4}$$
$$frac{sin alpha}{sin(180^circ-beta)} = frac{sin 20^circ}{sin 80^circ} tag{5}$$
Now, $alpha = 20^circ$ and $beta = 100^circ$, but when I plug these values in $triangle ABC$, it's not even triangle. oO
Where I am wrong? Thanks.
PS : sorry for poor editing, I don't have any clue about it.
geometry
edited Jan 19 at 9:38
Blue
48.6k870156
asked Jan 19 at 9:22
Angelus MortisAngelus Mortis
1356
$endgroup$
|
show 11 more comments
$begingroup$
In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, point $D$ is on $AC$, with $BC=AD$. Find $angle DBC$.
I know the correct solution, but I'm more interested in where is the problem in my solution.
My solution :
Now in $triangle ABD$, applying the sine rule:
$$frac{AD}{sinalpha} = frac{BD}{sin 20^circ} tag{1}$$
In $triangle BDC$:
$$frac{BD}{sin 80^circ} = frac{BC}{sin(180^circ-beta)} tag{2}$$
We know $AD= BC$; put in to $(1)$:
$$frac{BC}{sinalpha} = frac{BD}{sin 20^circ} tag{3}$$
Comparing $(2)$ and $(3)$:
$$frac{BC}{BD} = frac{sinalpha}{sin 20^circ} = frac{sin(180^circ-beta)}{sin 80^circ} tag{4}$$
$$frac{sin alpha}{sin(180^circ-beta)} = frac{sin 20^circ}{sin 80^circ} tag{5}$$
Now, $alpha = 20^circ$ and $beta = 100^circ$, but when I plug these values in $triangle ABC$, it's not even triangle. oO
Where I am wrong? Thanks.
PS : sorry for poor editing, I don't have any clue about it.
geometry
edited Jan 19 at 9:38
Blue
48.6k870156
asked Jan 19 at 9:22
Angelus MortisAngelus Mortis
1356
$endgroup$
1
$begingroup$
Full marks for the diagram, but please visit the MathJax documentation page and shore up the above by yourself. Note that MathJax is much easier to read, and a question that is good to read gets more attention, so fifteen minutes of learning the basics gets you a lot of attention on your questions and better answers.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:27
1
$begingroup$
@AngelusMortis: I'm making some edits to show you how it's done. Please don't edit until I'm finished. :)
$endgroup$
– Blue
Jan 19 at 9:31
1
$begingroup$
Good to see you taking the message on board. As for the actual question, from $frac{sin(alpha)}{sin(180-beta)} = frac{sin 20}{sin 80}$, how did you get to $alpha = 20$ and $beta = 100$? That part you have not explained. Ok, I see : you just compared numerators and denominators, and just took $alpha = 20$ and $180-beta = 80$ so $beta = 100$. That must be the incorrect step here, then.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:31
2
$begingroup$
I see. But that does not work here, unfortunately. Instead, note that $sin (180 - beta) = sin beta$ from the sum of sines formula, and also that $beta + alpha = 160$ so $160 - alpha = beta$. So substituting, gets you $frac{sin alpha}{sin(160-alpha)} = frac{sin 20}{sin 80}$. See what you can do from here. Also, since you know the correct value of $alpha$, check that it satisfies the above equation.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:37
1
$begingroup$
@Blue Thanks for helping with editing. Much appreciation sir :)
$endgroup$
– Angelus Mortis
Jan 19 at 10:59
|
show 11 more comments
$begingroup$
In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, point $D$ is on $AC$, with $BC=AD$. Find $angle DBC$.
I know the correct solution, but I'm more interested in where is the problem in my solution.
My solution :
Now in $triangle ABD$, applying the sine rule:
$$frac{AD}{sinalpha} = frac{BD}{sin 20^circ} tag{1}$$
In $triangle BDC$:
$$frac{BD}{sin 80^circ} = frac{BC}{sin(180^circ-beta)} tag{2}$$
We know $AD= BC$; put in to $(1)$:
$$frac{BC}{sinalpha} = frac{BD}{sin 20^circ} tag{3}$$
Comparing $(2)$ and $(3)$:
$$frac{BC}{BD} = frac{sinalpha}{sin 20^circ} = frac{sin(180^circ-beta)}{sin 80^circ} tag{4}$$
$$frac{sin alpha}{sin(180^circ-beta)} = frac{sin 20^circ}{sin 80^circ} tag{5}$$
Now, $alpha = 20^circ$ and $beta = 100^circ$, but when I plug these values in $triangle ABC$, it's not even triangle. oO
Where I am wrong? Thanks.
PS : sorry for poor editing, I don't have any clue about it.
geometry
edited Jan 19 at 9:38
Blue
48.6k870156
asked Jan 19 at 9:22
Angelus MortisAngelus Mortis
1356
$endgroup$
In $triangle ABC$ with $AB=AC$ and $angle BAC=20^circ$, point $D$ is on $AC$, with $BC=AD$. Find $angle DBC$.
I know the correct solution, but I'm more interested in where is the problem in my solution.
My solution :
Now in $triangle ABD$, applying the sine rule:
$$frac{AD}{sinalpha} = frac{BD}{sin 20^circ} tag{1}$$
In $triangle BDC$:
$$frac{BD}{sin 80^circ} = frac{BC}{sin(180^circ-beta)} tag{2}$$
We know $AD= BC$; put in to $(1)$:
$$frac{BC}{sinalpha} = frac{BD}{sin 20^circ} tag{3}$$
Comparing $(2)$ and $(3)$:
$$frac{BC}{BD} = frac{sinalpha}{sin 20^circ} = frac{sin(180^circ-beta)}{sin 80^circ} tag{4}$$
$$frac{sin alpha}{sin(180^circ-beta)} = frac{sin 20^circ}{sin 80^circ} tag{5}$$
Now, $alpha = 20^circ$ and $beta = 100^circ$, but when I plug these values in $triangle ABC$, it's not even triangle. oO
Where I am wrong? Thanks.
PS : sorry for poor editing, I don't have any clue about it.
geometry
geometry
edited Jan 19 at 9:38
Blue
48.6k870156
asked Jan 19 at 9:22
Angelus MortisAngelus Mortis
1356
edited Jan 19 at 9:38
Blue
48.6k870156
asked Jan 19 at 9:22
Angelus MortisAngelus Mortis
1356
edited Jan 19 at 9:38
Blue
48.6k870156
edited Jan 19 at 9:38
Blue
48.6k870156
edited Jan 19 at 9:38
Blue
48.6k870156
48.6k870156
asked Jan 19 at 9:22
Angelus MortisAngelus Mortis
1356
asked Jan 19 at 9:22
Angelus MortisAngelus Mortis
1356
asked Jan 19 at 9:22
Angelus MortisAngelus Mortis
1356
1356
1
$begingroup$
Full marks for the diagram, but please visit the MathJax documentation page and shore up the above by yourself. Note that MathJax is much easier to read, and a question that is good to read gets more attention, so fifteen minutes of learning the basics gets you a lot of attention on your questions and better answers.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:27
1
$begingroup$
@AngelusMortis: I'm making some edits to show you how it's done. Please don't edit until I'm finished. :)
$endgroup$
– Blue
Jan 19 at 9:31
1
$begingroup$
Good to see you taking the message on board. As for the actual question, from $frac{sin(alpha)}{sin(180-beta)} = frac{sin 20}{sin 80}$, how did you get to $alpha = 20$ and $beta = 100$? That part you have not explained. Ok, I see : you just compared numerators and denominators, and just took $alpha = 20$ and $180-beta = 80$ so $beta = 100$. That must be the incorrect step here, then.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:31
2
$begingroup$
I see. But that does not work here, unfortunately. Instead, note that $sin (180 - beta) = sin beta$ from the sum of sines formula, and also that $beta + alpha = 160$ so $160 - alpha = beta$. So substituting, gets you $frac{sin alpha}{sin(160-alpha)} = frac{sin 20}{sin 80}$. See what you can do from here. Also, since you know the correct value of $alpha$, check that it satisfies the above equation.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:37
1
$begingroup$
@Blue Thanks for helping with editing. Much appreciation sir :)
$endgroup$
– Angelus Mortis
Jan 19 at 10:59
|
show 11 more comments
1
$begingroup$
Full marks for the diagram, but please visit the MathJax documentation page and shore up the above by yourself. Note that MathJax is much easier to read, and a question that is good to read gets more attention, so fifteen minutes of learning the basics gets you a lot of attention on your questions and better answers.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:27
1
$begingroup$
@AngelusMortis: I'm making some edits to show you how it's done. Please don't edit until I'm finished. :)
$endgroup$
– Blue
Jan 19 at 9:31
1
$begingroup$
Good to see you taking the message on board. As for the actual question, from $frac{sin(alpha)}{sin(180-beta)} = frac{sin 20}{sin 80}$, how did you get to $alpha = 20$ and $beta = 100$? That part you have not explained. Ok, I see : you just compared numerators and denominators, and just took $alpha = 20$ and $180-beta = 80$ so $beta = 100$. That must be the incorrect step here, then.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:31
2
$begingroup$
I see. But that does not work here, unfortunately. Instead, note that $sin (180 - beta) = sin beta$ from the sum of sines formula, and also that $beta + alpha = 160$ so $160 - alpha = beta$. So substituting, gets you $frac{sin alpha}{sin(160-alpha)} = frac{sin 20}{sin 80}$. See what you can do from here. Also, since you know the correct value of $alpha$, check that it satisfies the above equation.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:37
1
$begingroup$
@Blue Thanks for helping with editing. Much appreciation sir :)
$endgroup$
– Angelus Mortis
Jan 19 at 10:59
1
1
$begingroup$
Full marks for the diagram, but please visit the MathJax documentation page and shore up the above by yourself. Note that MathJax is much easier to read, and a question that is good to read gets more attention, so fifteen minutes of learning the basics gets you a lot of attention on your questions and better answers.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:27
$begingroup$
Full marks for the diagram, but please visit the MathJax documentation page and shore up the above by yourself. Note that MathJax is much easier to read, and a question that is good to read gets more attention, so fifteen minutes of learning the basics gets you a lot of attention on your questions and better answers.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:27
1
1
$begingroup$
@AngelusMortis: I'm making some edits to show you how it's done. Please don't edit until I'm finished. :)
$endgroup$
– Blue
Jan 19 at 9:31
$begingroup$
@AngelusMortis: I'm making some edits to show you how it's done. Please don't edit until I'm finished. :)
$endgroup$
– Blue
Jan 19 at 9:31
1
1
$begingroup$
Good to see you taking the message on board. As for the actual question, from $frac{sin(alpha)}{sin(180-beta)} = frac{sin 20}{sin 80}$, how did you get to $alpha = 20$ and $beta = 100$? That part you have not explained. Ok, I see : you just compared numerators and denominators, and just took $alpha = 20$ and $180-beta = 80$ so $beta = 100$. That must be the incorrect step here, then.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:31
$begingroup$
Good to see you taking the message on board. As for the actual question, from $frac{sin(alpha)}{sin(180-beta)} = frac{sin 20}{sin 80}$, how did you get to $alpha = 20$ and $beta = 100$? That part you have not explained. Ok, I see : you just compared numerators and denominators, and just took $alpha = 20$ and $180-beta = 80$ so $beta = 100$. That must be the incorrect step here, then.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:31
2
2
$begingroup$
I see. But that does not work here, unfortunately. Instead, note that $sin (180 - beta) = sin beta$ from the sum of sines formula, and also that $beta + alpha = 160$ so $160 - alpha = beta$. So substituting, gets you $frac{sin alpha}{sin(160-alpha)} = frac{sin 20}{sin 80}$. See what you can do from here. Also, since you know the correct value of $alpha$, check that it satisfies the above equation.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:37
$begingroup$
I see. But that does not work here, unfortunately. Instead, note that $sin (180 - beta) = sin beta$ from the sum of sines formula, and also that $beta + alpha = 160$ so $160 - alpha = beta$. So substituting, gets you $frac{sin alpha}{sin(160-alpha)} = frac{sin 20}{sin 80}$. See what you can do from here. Also, since you know the correct value of $alpha$, check that it satisfies the above equation.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:37
1
1
$begingroup$
@Blue Thanks for helping with editing. Much appreciation sir :)
$endgroup$
– Angelus Mortis
Jan 19 at 10:59
$begingroup$
@Blue Thanks for helping with editing. Much appreciation sir :)
$endgroup$
– Angelus Mortis
Jan 19 at 10:59
|
show 11 more comments
1 Answer
1
active
oldest
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$begingroup$
So we have that $frac{sin alpha}{sin (180-beta)} = frac{sin 20}{sin 80}$.
The first thing we use is that $alpha + beta = 160$ from the triangle $ABD$. From here, $180 - beta = 180 - (160-alpha) = 20 + alpha$.
Next, we note that:
$$
frac{sin 20}{sin 80} = frac{sin 20}{cos(90-80)} = frac{sin 20}{cos 10} = frac{2 sin 10 cos 10} {cos 10} = 2 sin 10
$$
So, we have the equation :
$$
frac{sin alpha}{sin (alpha + 20)} = 2 sin 10\ implies sin alpha = 2 sin 10 sin (20+alpha) = 2 sin 10 sin 20 cos alpha + 2 sin 10 cos 20 sin alpha
$$
Now, collecting terms of $sin alpha$ on one side, and facctorizing it out,
$$
sin alpha(1 - 2 sin 10 cos 20) = 2 sin 10 sin 20 cos alpha \ implies
tan alpha = frac{2 sin 10 sin 20}{1 - 2 sin 10 cos 20}
$$
The right hand side is some fixed number which we have to find.
To do this, we first simplify the denominator, using the formulas : $$2 sin Acos B = sin(A+B) + sin(A-B) quad ; quad sin A + sin B = 2 sinleft(frac{A+B}{2}right)cosleft(frac{A-B}{2}right)$$
We will also use the fact that $sin 30 = frac 12$.
In our case,
$$
1 - 2 sin 10 cos 20 = 1- (sin 30 + sin (-10)) = 2 sin 30 - (sin 30 - sin 10) \ = sin 30 +sin 10 = 2 sin 20 cos 10
$$
Therefore,
$$
tan alpha = frac{2 sin 10 sin 20}{1 - 2 sin 10 cos 20} = frac{2sin 10 sin 20}{2 cos 10 sin 20} = tan 10
$$
Now, since $0 < alpha < 180$, we get that $alpha = 10$. From here, $80-alpha = 70$ is the desired angle.
answered Jan 20 at 12:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
add a comment |
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$begingroup$
So we have that $frac{sin alpha}{sin (180-beta)} = frac{sin 20}{sin 80}$.
The first thing we use is that $alpha + beta = 160$ from the triangle $ABD$. From here, $180 - beta = 180 - (160-alpha) = 20 + alpha$.
Next, we note that:
$$
frac{sin 20}{sin 80} = frac{sin 20}{cos(90-80)} = frac{sin 20}{cos 10} = frac{2 sin 10 cos 10} {cos 10} = 2 sin 10
$$
So, we have the equation :
$$
frac{sin alpha}{sin (alpha + 20)} = 2 sin 10\ implies sin alpha = 2 sin 10 sin (20+alpha) = 2 sin 10 sin 20 cos alpha + 2 sin 10 cos 20 sin alpha
$$
Now, collecting terms of $sin alpha$ on one side, and facctorizing it out,
$$
sin alpha(1 - 2 sin 10 cos 20) = 2 sin 10 sin 20 cos alpha \ implies
tan alpha = frac{2 sin 10 sin 20}{1 - 2 sin 10 cos 20}
$$
The right hand side is some fixed number which we have to find.
To do this, we first simplify the denominator, using the formulas : $$2 sin Acos B = sin(A+B) + sin(A-B) quad ; quad sin A + sin B = 2 sinleft(frac{A+B}{2}right)cosleft(frac{A-B}{2}right)$$
We will also use the fact that $sin 30 = frac 12$.
In our case,
$$
1 - 2 sin 10 cos 20 = 1- (sin 30 + sin (-10)) = 2 sin 30 - (sin 30 - sin 10) \ = sin 30 +sin 10 = 2 sin 20 cos 10
$$
Therefore,
$$
tan alpha = frac{2 sin 10 sin 20}{1 - 2 sin 10 cos 20} = frac{2sin 10 sin 20}{2 cos 10 sin 20} = tan 10
$$
Now, since $0 < alpha < 180$, we get that $alpha = 10$. From here, $80-alpha = 70$ is the desired angle.
answered Jan 20 at 12:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
add a comment |
$begingroup$
So we have that $frac{sin alpha}{sin (180-beta)} = frac{sin 20}{sin 80}$.
The first thing we use is that $alpha + beta = 160$ from the triangle $ABD$. From here, $180 - beta = 180 - (160-alpha) = 20 + alpha$.
Next, we note that:
$$
frac{sin 20}{sin 80} = frac{sin 20}{cos(90-80)} = frac{sin 20}{cos 10} = frac{2 sin 10 cos 10} {cos 10} = 2 sin 10
$$
So, we have the equation :
$$
frac{sin alpha}{sin (alpha + 20)} = 2 sin 10\ implies sin alpha = 2 sin 10 sin (20+alpha) = 2 sin 10 sin 20 cos alpha + 2 sin 10 cos 20 sin alpha
$$
Now, collecting terms of $sin alpha$ on one side, and facctorizing it out,
$$
sin alpha(1 - 2 sin 10 cos 20) = 2 sin 10 sin 20 cos alpha \ implies
tan alpha = frac{2 sin 10 sin 20}{1 - 2 sin 10 cos 20}
$$
The right hand side is some fixed number which we have to find.
To do this, we first simplify the denominator, using the formulas : $$2 sin Acos B = sin(A+B) + sin(A-B) quad ; quad sin A + sin B = 2 sinleft(frac{A+B}{2}right)cosleft(frac{A-B}{2}right)$$
We will also use the fact that $sin 30 = frac 12$.
In our case,
$$
1 - 2 sin 10 cos 20 = 1- (sin 30 + sin (-10)) = 2 sin 30 - (sin 30 - sin 10) \ = sin 30 +sin 10 = 2 sin 20 cos 10
$$
Therefore,
$$
tan alpha = frac{2 sin 10 sin 20}{1 - 2 sin 10 cos 20} = frac{2sin 10 sin 20}{2 cos 10 sin 20} = tan 10
$$
Now, since $0 < alpha < 180$, we get that $alpha = 10$. From here, $80-alpha = 70$ is the desired angle.
answered Jan 20 at 12:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
add a comment |
$begingroup$
So we have that $frac{sin alpha}{sin (180-beta)} = frac{sin 20}{sin 80}$.
The first thing we use is that $alpha + beta = 160$ from the triangle $ABD$. From here, $180 - beta = 180 - (160-alpha) = 20 + alpha$.
Next, we note that:
$$
frac{sin 20}{sin 80} = frac{sin 20}{cos(90-80)} = frac{sin 20}{cos 10} = frac{2 sin 10 cos 10} {cos 10} = 2 sin 10
$$
So, we have the equation :
$$
frac{sin alpha}{sin (alpha + 20)} = 2 sin 10\ implies sin alpha = 2 sin 10 sin (20+alpha) = 2 sin 10 sin 20 cos alpha + 2 sin 10 cos 20 sin alpha
$$
Now, collecting terms of $sin alpha$ on one side, and facctorizing it out,
$$
sin alpha(1 - 2 sin 10 cos 20) = 2 sin 10 sin 20 cos alpha \ implies
tan alpha = frac{2 sin 10 sin 20}{1 - 2 sin 10 cos 20}
$$
The right hand side is some fixed number which we have to find.
To do this, we first simplify the denominator, using the formulas : $$2 sin Acos B = sin(A+B) + sin(A-B) quad ; quad sin A + sin B = 2 sinleft(frac{A+B}{2}right)cosleft(frac{A-B}{2}right)$$
We will also use the fact that $sin 30 = frac 12$.
In our case,
$$
1 - 2 sin 10 cos 20 = 1- (sin 30 + sin (-10)) = 2 sin 30 - (sin 30 - sin 10) \ = sin 30 +sin 10 = 2 sin 20 cos 10
$$
Therefore,
$$
tan alpha = frac{2 sin 10 sin 20}{1 - 2 sin 10 cos 20} = frac{2sin 10 sin 20}{2 cos 10 sin 20} = tan 10
$$
Now, since $0 < alpha < 180$, we get that $alpha = 10$. From here, $80-alpha = 70$ is the desired angle.
answered Jan 20 at 12:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
So we have that $frac{sin alpha}{sin (180-beta)} = frac{sin 20}{sin 80}$.
The first thing we use is that $alpha + beta = 160$ from the triangle $ABD$. From here, $180 - beta = 180 - (160-alpha) = 20 + alpha$.
Next, we note that:
$$
frac{sin 20}{sin 80} = frac{sin 20}{cos(90-80)} = frac{sin 20}{cos 10} = frac{2 sin 10 cos 10} {cos 10} = 2 sin 10
$$
So, we have the equation :
$$
frac{sin alpha}{sin (alpha + 20)} = 2 sin 10\ implies sin alpha = 2 sin 10 sin (20+alpha) = 2 sin 10 sin 20 cos alpha + 2 sin 10 cos 20 sin alpha
$$
Now, collecting terms of $sin alpha$ on one side, and facctorizing it out,
$$
sin alpha(1 - 2 sin 10 cos 20) = 2 sin 10 sin 20 cos alpha \ implies
tan alpha = frac{2 sin 10 sin 20}{1 - 2 sin 10 cos 20}
$$
The right hand side is some fixed number which we have to find.
To do this, we first simplify the denominator, using the formulas : $$2 sin Acos B = sin(A+B) + sin(A-B) quad ; quad sin A + sin B = 2 sinleft(frac{A+B}{2}right)cosleft(frac{A-B}{2}right)$$
We will also use the fact that $sin 30 = frac 12$.
In our case,
$$
1 - 2 sin 10 cos 20 = 1- (sin 30 + sin (-10)) = 2 sin 30 - (sin 30 - sin 10) \ = sin 30 +sin 10 = 2 sin 20 cos 10
$$
Therefore,
$$
tan alpha = frac{2 sin 10 sin 20}{1 - 2 sin 10 cos 20} = frac{2sin 10 sin 20}{2 cos 10 sin 20} = tan 10
$$
Now, since $0 < alpha < 180$, we get that $alpha = 10$. From here, $80-alpha = 70$ is the desired angle.
answered Jan 20 at 12:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 20 at 12:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 20 at 12:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 20 at 12:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
38.9k33477
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1
$begingroup$
Full marks for the diagram, but please visit the MathJax documentation page and shore up the above by yourself. Note that MathJax is much easier to read, and a question that is good to read gets more attention, so fifteen minutes of learning the basics gets you a lot of attention on your questions and better answers.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:27
1
$begingroup$
@AngelusMortis: I'm making some edits to show you how it's done. Please don't edit until I'm finished. :)
$endgroup$
– Blue
Jan 19 at 9:31
1
$begingroup$
Good to see you taking the message on board. As for the actual question, from $frac{sin(alpha)}{sin(180-beta)} = frac{sin 20}{sin 80}$, how did you get to $alpha = 20$ and $beta = 100$? That part you have not explained. Ok, I see : you just compared numerators and denominators, and just took $alpha = 20$ and $180-beta = 80$ so $beta = 100$. That must be the incorrect step here, then.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:31
2
$begingroup$
I see. But that does not work here, unfortunately. Instead, note that $sin (180 - beta) = sin beta$ from the sum of sines formula, and also that $beta + alpha = 160$ so $160 - alpha = beta$. So substituting, gets you $frac{sin alpha}{sin(160-alpha)} = frac{sin 20}{sin 80}$. See what you can do from here. Also, since you know the correct value of $alpha$, check that it satisfies the above equation.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 9:37
1
$begingroup$
@Blue Thanks for helping with editing. Much appreciation sir :)
$endgroup$
– Angelus Mortis
Jan 19 at 10:59