Mixing
Cannot find angle for trigonometry problem
$begingroup$
A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $frac{13}{4}$. What are the angles of the triangle?
Here is the problem as I understand it, given that:
1. $measuredangle ABC = 90^circ rightarrow AC = 2R$
2. The smaller radii are perpendicular to the triangle.
3. The origin of the smaller circle is the intersection between the angle bisectors of triangle ABC.
Let $measuredangle ACB = 2alpha rightarrow measuredangle ACG = alpha$
$measuredangle BAC = 90 - 2alpha rightarrow measuredangle GAC = 45 - alpha$
From here we can construct the following system:
$$begin{cases} 2R = frac{r}{tan alpha} + frac{r}{tan(45 - alpha)} \ frac{R}{r} = frac{13}{4} rightarrow R = frac{13r}{4} end{cases}$$
$$downarrow \ frac{26r}{4} = rleft(frac{1}{tan alpha} + frac{1}{tan(45 - alpha)}right) quad / div r \ 6.5 = frac{cos alpha}{sin alpha} + frac{cos(45 - alpha)}{sin(45 - alpha)} quad / {sin(alpha pm beta) = sinalphacosbeta pm sinbetacosalpha\ cos(alpha pm beta) = cosalphacosbeta mp sinalphasinbeta} \ 6.5 = frac{cos alpha}{sin alpha} + frac{cos alpha cos 45 + sin alpha sin 45}{sin 45 cos alpha - sin alpha cos 45} \ 6.5 = frac{cos alpha}{sin alpha } + frac{frac{sqrt{2}}{2}}{frac{sqrt{2}}{2}} cdot frac{cos alpha + sin alpha}{cos alpha - sin alpha} \ 6.5 = frac{cos^2 alpha - sinalpha cos alpha + cos alpha sin alpha + sin^2 alpha}{sin(cosalpha - sin alpha)} quad / sin^2alpha + cos ^2 alpha = 1\ 6.5 = frac{1}{sinalpha (cosalpha - sin alpha)}$$
From this point on, I have no identity that I can think of which can be applied here to help me solve for $alpha$. Amusingly enough, I got an answer which was very close to the correct one, but that was only because I made a mistake in the signs of the identity, and the second fracture turned into 1.
I am aware that they may be an identity which can solve this, but I am restricted to only use the identities available here. I have either made a mistake somewhere, or just don't know how to use the available identities to solve the problem.
geometry trigonometry
asked Jan 23 at 18:04
daedsidogdaedsidog
29517
$endgroup$
add a comment |
$begingroup$
A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $frac{13}{4}$. What are the angles of the triangle?
Here is the problem as I understand it, given that:
1. $measuredangle ABC = 90^circ rightarrow AC = 2R$
2. The smaller radii are perpendicular to the triangle.
3. The origin of the smaller circle is the intersection between the angle bisectors of triangle ABC.
Let $measuredangle ACB = 2alpha rightarrow measuredangle ACG = alpha$
$measuredangle BAC = 90 - 2alpha rightarrow measuredangle GAC = 45 - alpha$
From here we can construct the following system:
$$begin{cases} 2R = frac{r}{tan alpha} + frac{r}{tan(45 - alpha)} \ frac{R}{r} = frac{13}{4} rightarrow R = frac{13r}{4} end{cases}$$
$$downarrow \ frac{26r}{4} = rleft(frac{1}{tan alpha} + frac{1}{tan(45 - alpha)}right) quad / div r \ 6.5 = frac{cos alpha}{sin alpha} + frac{cos(45 - alpha)}{sin(45 - alpha)} quad / {sin(alpha pm beta) = sinalphacosbeta pm sinbetacosalpha\ cos(alpha pm beta) = cosalphacosbeta mp sinalphasinbeta} \ 6.5 = frac{cos alpha}{sin alpha} + frac{cos alpha cos 45 + sin alpha sin 45}{sin 45 cos alpha - sin alpha cos 45} \ 6.5 = frac{cos alpha}{sin alpha } + frac{frac{sqrt{2}}{2}}{frac{sqrt{2}}{2}} cdot frac{cos alpha + sin alpha}{cos alpha - sin alpha} \ 6.5 = frac{cos^2 alpha - sinalpha cos alpha + cos alpha sin alpha + sin^2 alpha}{sin(cosalpha - sin alpha)} quad / sin^2alpha + cos ^2 alpha = 1\ 6.5 = frac{1}{sinalpha (cosalpha - sin alpha)}$$
From this point on, I have no identity that I can think of which can be applied here to help me solve for $alpha$. Amusingly enough, I got an answer which was very close to the correct one, but that was only because I made a mistake in the signs of the identity, and the second fracture turned into 1.
I am aware that they may be an identity which can solve this, but I am restricted to only use the identities available here. I have either made a mistake somewhere, or just don't know how to use the available identities to solve the problem.
geometry trigonometry
asked Jan 23 at 18:04
daedsidogdaedsidog
29517
$endgroup$
$begingroup$
try rewriting your last equation as $cosalpha = dfrac{1}{6.5sinalpha}+sinalpha$. What happens when you square both sides?
$endgroup$
– John Joy
Jan 23 at 18:37
$begingroup$
@JohnJoy I worked it out your way, and I get two possible solutions for an angle, with one of them being correct after disproving the other. However, I think that this amount of calculation is much beyond the level of this problem, and I suspect there's a simpler solution. Nevertheless, this works very well — thank you.
$endgroup$
– daedsidog
Jan 23 at 18:58
add a comment |
$begingroup$
A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $frac{13}{4}$. What are the angles of the triangle?
Here is the problem as I understand it, given that:
1. $measuredangle ABC = 90^circ rightarrow AC = 2R$
2. The smaller radii are perpendicular to the triangle.
3. The origin of the smaller circle is the intersection between the angle bisectors of triangle ABC.
Let $measuredangle ACB = 2alpha rightarrow measuredangle ACG = alpha$
$measuredangle BAC = 90 - 2alpha rightarrow measuredangle GAC = 45 - alpha$
From here we can construct the following system:
$$begin{cases} 2R = frac{r}{tan alpha} + frac{r}{tan(45 - alpha)} \ frac{R}{r} = frac{13}{4} rightarrow R = frac{13r}{4} end{cases}$$
$$downarrow \ frac{26r}{4} = rleft(frac{1}{tan alpha} + frac{1}{tan(45 - alpha)}right) quad / div r \ 6.5 = frac{cos alpha}{sin alpha} + frac{cos(45 - alpha)}{sin(45 - alpha)} quad / {sin(alpha pm beta) = sinalphacosbeta pm sinbetacosalpha\ cos(alpha pm beta) = cosalphacosbeta mp sinalphasinbeta} \ 6.5 = frac{cos alpha}{sin alpha} + frac{cos alpha cos 45 + sin alpha sin 45}{sin 45 cos alpha - sin alpha cos 45} \ 6.5 = frac{cos alpha}{sin alpha } + frac{frac{sqrt{2}}{2}}{frac{sqrt{2}}{2}} cdot frac{cos alpha + sin alpha}{cos alpha - sin alpha} \ 6.5 = frac{cos^2 alpha - sinalpha cos alpha + cos alpha sin alpha + sin^2 alpha}{sin(cosalpha - sin alpha)} quad / sin^2alpha + cos ^2 alpha = 1\ 6.5 = frac{1}{sinalpha (cosalpha - sin alpha)}$$
From this point on, I have no identity that I can think of which can be applied here to help me solve for $alpha$. Amusingly enough, I got an answer which was very close to the correct one, but that was only because I made a mistake in the signs of the identity, and the second fracture turned into 1.
I am aware that they may be an identity which can solve this, but I am restricted to only use the identities available here. I have either made a mistake somewhere, or just don't know how to use the available identities to solve the problem.
geometry trigonometry
asked Jan 23 at 18:04
daedsidogdaedsidog
29517
$endgroup$
A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $frac{13}{4}$. What are the angles of the triangle?
Here is the problem as I understand it, given that:
1. $measuredangle ABC = 90^circ rightarrow AC = 2R$
2. The smaller radii are perpendicular to the triangle.
3. The origin of the smaller circle is the intersection between the angle bisectors of triangle ABC.
Let $measuredangle ACB = 2alpha rightarrow measuredangle ACG = alpha$
$measuredangle BAC = 90 - 2alpha rightarrow measuredangle GAC = 45 - alpha$
From here we can construct the following system:
$$begin{cases} 2R = frac{r}{tan alpha} + frac{r}{tan(45 - alpha)} \ frac{R}{r} = frac{13}{4} rightarrow R = frac{13r}{4} end{cases}$$
$$downarrow \ frac{26r}{4} = rleft(frac{1}{tan alpha} + frac{1}{tan(45 - alpha)}right) quad / div r \ 6.5 = frac{cos alpha}{sin alpha} + frac{cos(45 - alpha)}{sin(45 - alpha)} quad / {sin(alpha pm beta) = sinalphacosbeta pm sinbetacosalpha\ cos(alpha pm beta) = cosalphacosbeta mp sinalphasinbeta} \ 6.5 = frac{cos alpha}{sin alpha} + frac{cos alpha cos 45 + sin alpha sin 45}{sin 45 cos alpha - sin alpha cos 45} \ 6.5 = frac{cos alpha}{sin alpha } + frac{frac{sqrt{2}}{2}}{frac{sqrt{2}}{2}} cdot frac{cos alpha + sin alpha}{cos alpha - sin alpha} \ 6.5 = frac{cos^2 alpha - sinalpha cos alpha + cos alpha sin alpha + sin^2 alpha}{sin(cosalpha - sin alpha)} quad / sin^2alpha + cos ^2 alpha = 1\ 6.5 = frac{1}{sinalpha (cosalpha - sin alpha)}$$
From this point on, I have no identity that I can think of which can be applied here to help me solve for $alpha$. Amusingly enough, I got an answer which was very close to the correct one, but that was only because I made a mistake in the signs of the identity, and the second fracture turned into 1.
I am aware that they may be an identity which can solve this, but I am restricted to only use the identities available here. I have either made a mistake somewhere, or just don't know how to use the available identities to solve the problem.
geometry trigonometry
geometry trigonometry
asked Jan 23 at 18:04
daedsidogdaedsidog
29517
asked Jan 23 at 18:04
daedsidogdaedsidog
29517
edited Jan 23 at 18:09
daedsidog
asked Jan 23 at 18:04
daedsidogdaedsidog
29517
asked Jan 23 at 18:04
daedsidogdaedsidog
29517
asked Jan 23 at 18:04
daedsidogdaedsidog
29517
29517
$begingroup$
try rewriting your last equation as $cosalpha = dfrac{1}{6.5sinalpha}+sinalpha$. What happens when you square both sides?
$endgroup$
– John Joy
Jan 23 at 18:37
$begingroup$
@JohnJoy I worked it out your way, and I get two possible solutions for an angle, with one of them being correct after disproving the other. However, I think that this amount of calculation is much beyond the level of this problem, and I suspect there's a simpler solution. Nevertheless, this works very well — thank you.
$endgroup$
– daedsidog
Jan 23 at 18:58
add a comment |
$begingroup$
try rewriting your last equation as $cosalpha = dfrac{1}{6.5sinalpha}+sinalpha$. What happens when you square both sides?
$endgroup$
– John Joy
Jan 23 at 18:37
$begingroup$
@JohnJoy I worked it out your way, and I get two possible solutions for an angle, with one of them being correct after disproving the other. However, I think that this amount of calculation is much beyond the level of this problem, and I suspect there's a simpler solution. Nevertheless, this works very well — thank you.
$endgroup$
– daedsidog
Jan 23 at 18:58
$begingroup$
try rewriting your last equation as $cosalpha = dfrac{1}{6.5sinalpha}+sinalpha$. What happens when you square both sides?
$endgroup$
– John Joy
Jan 23 at 18:37
$begingroup$
try rewriting your last equation as $cosalpha = dfrac{1}{6.5sinalpha}+sinalpha$. What happens when you square both sides?
$endgroup$
– John Joy
Jan 23 at 18:37
$begingroup$
@JohnJoy I worked it out your way, and I get two possible solutions for an angle, with one of them being correct after disproving the other. However, I think that this amount of calculation is much beyond the level of this problem, and I suspect there's a simpler solution. Nevertheless, this works very well — thank you.
$endgroup$
– daedsidog
Jan 23 at 18:58
$begingroup$
@JohnJoy I worked it out your way, and I get two possible solutions for an angle, with one of them being correct after disproving the other. However, I think that this amount of calculation is much beyond the level of this problem, and I suspect there's a simpler solution. Nevertheless, this works very well — thank you.
$endgroup$
– daedsidog
Jan 23 at 18:58
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We will use that$$2r=a+c-b$$ and $$2R=b$$ then $$16b=13a+13c-13b$$ so $$29b=13(a+c)$$ and with $$a^2+c^2=b^2$$ we can eliminate $b$ so we get
$$frac{a}{c}+frac{c}{a}=frac{169}{336}$$ from here you will get the quotient of $$frac {a}{c}$$ and thus we can calculate the angles of triangle $$Delta ABC$$
answered Jan 23 at 18:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
$begingroup$
What are $a$, $b$, and $c$? They certainly aren't the lengths of the triangle in your example.
$endgroup$
– daedsidog
Jan 23 at 19:10
$begingroup$
Why not? They are the side lengths of the given triangle
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
I have all written what you need to solve the problem
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
You need only three formulas: $$b^2=a^2+c^2,b=2R,a+c-b=2r$$ and the given quotient $$frac{R}{r}=frac{13}{4}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:13
$begingroup$
$2b neq R$, but $b = 2R$. Typo on your part?
$endgroup$
– daedsidog
Jan 23 at 19:16
|
show 3 more comments
$begingroup$
We have
$$2R=frac{r}{tan(alpha/2)}+frac{r}{tan(beta/2)},$$
hence
$$frac{13}{2}=frac{1}{tan(alpha/2)}+frac{1}{tan(beta/2)}.$$
We know that $alpha+beta=pi/2$, so $beta/2=pi/4-alpha/2$. Let $t=tan(alpha/2)$. From the addition theorems for $sin$ and $cos$ derive that
$$tan(pi/4-alpha/2)=frac{1-t}{1+t}.$$ Substitute and solve the quadratic equation for $t$.
answered Jan 23 at 19:25
Michael HoppeMichael Hoppe
11.2k31837
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add a comment |
$begingroup$
$$dfrac4{13}sin(45^circ -alpha+alpha)=2sinalphasin(45^circ-alpha)$$
$$=cos(45^circ-2alpha)-cos45^circ$$
answered Jan 23 at 18:15
lab bhattacharjeelab bhattacharjee
226k15158275
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$begingroup$
See mathworld.wolfram.com/WernerFormulas.html
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– lab bhattacharjee
Jan 23 at 18:16
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I cannot use those formulae.
$endgroup$
– daedsidog
Jan 23 at 18:19
$begingroup$
@daedsidog, can you use $cos(x-y)-cos(x+y)=?$
$endgroup$
– lab bhattacharjee
Jan 23 at 19:03
$begingroup$
Here's the list of identities I can use: roygeva.co.il/file/CKFiles/trig.pdf
$endgroup$
– daedsidog
Jan 23 at 19:05
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We will use that$$2r=a+c-b$$ and $$2R=b$$ then $$16b=13a+13c-13b$$ so $$29b=13(a+c)$$ and with $$a^2+c^2=b^2$$ we can eliminate $b$ so we get
$$frac{a}{c}+frac{c}{a}=frac{169}{336}$$ from here you will get the quotient of $$frac {a}{c}$$ and thus we can calculate the angles of triangle $$Delta ABC$$
answered Jan 23 at 18:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
$begingroup$
What are $a$, $b$, and $c$? They certainly aren't the lengths of the triangle in your example.
$endgroup$
– daedsidog
Jan 23 at 19:10
$begingroup$
Why not? They are the side lengths of the given triangle
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
I have all written what you need to solve the problem
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
You need only three formulas: $$b^2=a^2+c^2,b=2R,a+c-b=2r$$ and the given quotient $$frac{R}{r}=frac{13}{4}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:13
$begingroup$
$2b neq R$, but $b = 2R$. Typo on your part?
$endgroup$
– daedsidog
Jan 23 at 19:16
|
show 3 more comments
$begingroup$
We will use that$$2r=a+c-b$$ and $$2R=b$$ then $$16b=13a+13c-13b$$ so $$29b=13(a+c)$$ and with $$a^2+c^2=b^2$$ we can eliminate $b$ so we get
$$frac{a}{c}+frac{c}{a}=frac{169}{336}$$ from here you will get the quotient of $$frac {a}{c}$$ and thus we can calculate the angles of triangle $$Delta ABC$$
answered Jan 23 at 18:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
$begingroup$
What are $a$, $b$, and $c$? They certainly aren't the lengths of the triangle in your example.
$endgroup$
– daedsidog
Jan 23 at 19:10
$begingroup$
Why not? They are the side lengths of the given triangle
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
I have all written what you need to solve the problem
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
You need only three formulas: $$b^2=a^2+c^2,b=2R,a+c-b=2r$$ and the given quotient $$frac{R}{r}=frac{13}{4}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:13
$begingroup$
$2b neq R$, but $b = 2R$. Typo on your part?
$endgroup$
– daedsidog
Jan 23 at 19:16
|
show 3 more comments
$begingroup$
We will use that$$2r=a+c-b$$ and $$2R=b$$ then $$16b=13a+13c-13b$$ so $$29b=13(a+c)$$ and with $$a^2+c^2=b^2$$ we can eliminate $b$ so we get
$$frac{a}{c}+frac{c}{a}=frac{169}{336}$$ from here you will get the quotient of $$frac {a}{c}$$ and thus we can calculate the angles of triangle $$Delta ABC$$
answered Jan 23 at 18:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
$endgroup$
We will use that$$2r=a+c-b$$ and $$2R=b$$ then $$16b=13a+13c-13b$$ so $$29b=13(a+c)$$ and with $$a^2+c^2=b^2$$ we can eliminate $b$ so we get
$$frac{a}{c}+frac{c}{a}=frac{169}{336}$$ from here you will get the quotient of $$frac {a}{c}$$ and thus we can calculate the angles of triangle $$Delta ABC$$
answered Jan 23 at 18:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
edited Jan 23 at 19:18
answered Jan 23 at 18:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered Jan 23 at 18:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
answered Jan 23 at 18:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.7k42866
77.7k42866
$begingroup$
What are $a$, $b$, and $c$? They certainly aren't the lengths of the triangle in your example.
$endgroup$
– daedsidog
Jan 23 at 19:10
$begingroup$
Why not? They are the side lengths of the given triangle
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
I have all written what you need to solve the problem
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
You need only three formulas: $$b^2=a^2+c^2,b=2R,a+c-b=2r$$ and the given quotient $$frac{R}{r}=frac{13}{4}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:13
$begingroup$
$2b neq R$, but $b = 2R$. Typo on your part?
$endgroup$
– daedsidog
Jan 23 at 19:16
|
show 3 more comments
$begingroup$
What are $a$, $b$, and $c$? They certainly aren't the lengths of the triangle in your example.
$endgroup$
– daedsidog
Jan 23 at 19:10
$begingroup$
Why not? They are the side lengths of the given triangle
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
I have all written what you need to solve the problem
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
You need only three formulas: $$b^2=a^2+c^2,b=2R,a+c-b=2r$$ and the given quotient $$frac{R}{r}=frac{13}{4}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:13
$begingroup$
$2b neq R$, but $b = 2R$. Typo on your part?
$endgroup$
– daedsidog
Jan 23 at 19:16
$begingroup$
What are $a$, $b$, and $c$? They certainly aren't the lengths of the triangle in your example.
$endgroup$
– daedsidog
Jan 23 at 19:10
$begingroup$
What are $a$, $b$, and $c$? They certainly aren't the lengths of the triangle in your example.
$endgroup$
– daedsidog
Jan 23 at 19:10
$begingroup$
Why not? They are the side lengths of the given triangle
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
Why not? They are the side lengths of the given triangle
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
I have all written what you need to solve the problem
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
I have all written what you need to solve the problem
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:11
$begingroup$
You need only three formulas: $$b^2=a^2+c^2,b=2R,a+c-b=2r$$ and the given quotient $$frac{R}{r}=frac{13}{4}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:13
$begingroup$
You need only three formulas: $$b^2=a^2+c^2,b=2R,a+c-b=2r$$ and the given quotient $$frac{R}{r}=frac{13}{4}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 23 at 19:13
$begingroup$
$2b neq R$, but $b = 2R$. Typo on your part?
$endgroup$
– daedsidog
Jan 23 at 19:16
$begingroup$
$2b neq R$, but $b = 2R$. Typo on your part?
$endgroup$
– daedsidog
Jan 23 at 19:16
|
show 3 more comments
$begingroup$
We have
$$2R=frac{r}{tan(alpha/2)}+frac{r}{tan(beta/2)},$$
hence
$$frac{13}{2}=frac{1}{tan(alpha/2)}+frac{1}{tan(beta/2)}.$$
We know that $alpha+beta=pi/2$, so $beta/2=pi/4-alpha/2$. Let $t=tan(alpha/2)$. From the addition theorems for $sin$ and $cos$ derive that
$$tan(pi/4-alpha/2)=frac{1-t}{1+t}.$$ Substitute and solve the quadratic equation for $t$.
answered Jan 23 at 19:25
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
add a comment |
$begingroup$
We have
$$2R=frac{r}{tan(alpha/2)}+frac{r}{tan(beta/2)},$$
hence
$$frac{13}{2}=frac{1}{tan(alpha/2)}+frac{1}{tan(beta/2)}.$$
We know that $alpha+beta=pi/2$, so $beta/2=pi/4-alpha/2$. Let $t=tan(alpha/2)$. From the addition theorems for $sin$ and $cos$ derive that
$$tan(pi/4-alpha/2)=frac{1-t}{1+t}.$$ Substitute and solve the quadratic equation for $t$.
answered Jan 23 at 19:25
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
add a comment |
$begingroup$
We have
$$2R=frac{r}{tan(alpha/2)}+frac{r}{tan(beta/2)},$$
hence
$$frac{13}{2}=frac{1}{tan(alpha/2)}+frac{1}{tan(beta/2)}.$$
We know that $alpha+beta=pi/2$, so $beta/2=pi/4-alpha/2$. Let $t=tan(alpha/2)$. From the addition theorems for $sin$ and $cos$ derive that
$$tan(pi/4-alpha/2)=frac{1-t}{1+t}.$$ Substitute and solve the quadratic equation for $t$.
answered Jan 23 at 19:25
Michael HoppeMichael Hoppe
11.2k31837
$endgroup$
We have
$$2R=frac{r}{tan(alpha/2)}+frac{r}{tan(beta/2)},$$
hence
$$frac{13}{2}=frac{1}{tan(alpha/2)}+frac{1}{tan(beta/2)}.$$
We know that $alpha+beta=pi/2$, so $beta/2=pi/4-alpha/2$. Let $t=tan(alpha/2)$. From the addition theorems for $sin$ and $cos$ derive that
$$tan(pi/4-alpha/2)=frac{1-t}{1+t}.$$ Substitute and solve the quadratic equation for $t$.
answered Jan 23 at 19:25
Michael HoppeMichael Hoppe
11.2k31837
answered Jan 23 at 19:25
Michael HoppeMichael Hoppe
11.2k31837
answered Jan 23 at 19:25
Michael HoppeMichael Hoppe
11.2k31837
answered Jan 23 at 19:25
Michael HoppeMichael Hoppe
11.2k31837
11.2k31837
add a comment |
add a comment |
$begingroup$
$$dfrac4{13}sin(45^circ -alpha+alpha)=2sinalphasin(45^circ-alpha)$$
$$=cos(45^circ-2alpha)-cos45^circ$$
answered Jan 23 at 18:15
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
See mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Jan 23 at 18:16
$begingroup$
I cannot use those formulae.
$endgroup$
– daedsidog
Jan 23 at 18:19
$begingroup$
@daedsidog, can you use $cos(x-y)-cos(x+y)=?$
$endgroup$
– lab bhattacharjee
Jan 23 at 19:03
$begingroup$
Here's the list of identities I can use: roygeva.co.il/file/CKFiles/trig.pdf
$endgroup$
– daedsidog
Jan 23 at 19:05
add a comment |
$begingroup$
$$dfrac4{13}sin(45^circ -alpha+alpha)=2sinalphasin(45^circ-alpha)$$
$$=cos(45^circ-2alpha)-cos45^circ$$
answered Jan 23 at 18:15
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
See mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Jan 23 at 18:16
$begingroup$
I cannot use those formulae.
$endgroup$
– daedsidog
Jan 23 at 18:19
$begingroup$
@daedsidog, can you use $cos(x-y)-cos(x+y)=?$
$endgroup$
– lab bhattacharjee
Jan 23 at 19:03
$begingroup$
Here's the list of identities I can use: roygeva.co.il/file/CKFiles/trig.pdf
$endgroup$
– daedsidog
Jan 23 at 19:05
add a comment |
$begingroup$
$$dfrac4{13}sin(45^circ -alpha+alpha)=2sinalphasin(45^circ-alpha)$$
$$=cos(45^circ-2alpha)-cos45^circ$$
answered Jan 23 at 18:15
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$$dfrac4{13}sin(45^circ -alpha+alpha)=2sinalphasin(45^circ-alpha)$$
$$=cos(45^circ-2alpha)-cos45^circ$$
answered Jan 23 at 18:15
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 23 at 18:15
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 23 at 18:15
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 23 at 18:15
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
$begingroup$
See mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Jan 23 at 18:16
$begingroup$
I cannot use those formulae.
$endgroup$
– daedsidog
Jan 23 at 18:19
$begingroup$
@daedsidog, can you use $cos(x-y)-cos(x+y)=?$
$endgroup$
– lab bhattacharjee
Jan 23 at 19:03
$begingroup$
Here's the list of identities I can use: roygeva.co.il/file/CKFiles/trig.pdf
$endgroup$
– daedsidog
Jan 23 at 19:05
add a comment |
$begingroup$
See mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Jan 23 at 18:16
$begingroup$
I cannot use those formulae.
$endgroup$
– daedsidog
Jan 23 at 18:19
$begingroup$
@daedsidog, can you use $cos(x-y)-cos(x+y)=?$
$endgroup$
– lab bhattacharjee
Jan 23 at 19:03
$begingroup$
Here's the list of identities I can use: roygeva.co.il/file/CKFiles/trig.pdf
$endgroup$
– daedsidog
Jan 23 at 19:05
$begingroup$
See mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Jan 23 at 18:16
$begingroup$
See mathworld.wolfram.com/WernerFormulas.html
$endgroup$
– lab bhattacharjee
Jan 23 at 18:16
$begingroup$
I cannot use those formulae.
$endgroup$
– daedsidog
Jan 23 at 18:19
$begingroup$
I cannot use those formulae.
$endgroup$
– daedsidog
Jan 23 at 18:19
$begingroup$
@daedsidog, can you use $cos(x-y)-cos(x+y)=?$
$endgroup$
– lab bhattacharjee
Jan 23 at 19:03
$begingroup$
@daedsidog, can you use $cos(x-y)-cos(x+y)=?$
$endgroup$
– lab bhattacharjee
Jan 23 at 19:03
$begingroup$
Here's the list of identities I can use: roygeva.co.il/file/CKFiles/trig.pdf
$endgroup$
– daedsidog
Jan 23 at 19:05
$begingroup$
Here's the list of identities I can use: roygeva.co.il/file/CKFiles/trig.pdf
$endgroup$
– daedsidog
Jan 23 at 19:05
add a comment |
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$begingroup$
try rewriting your last equation as $cosalpha = dfrac{1}{6.5sinalpha}+sinalpha$. What happens when you square both sides?
$endgroup$
– John Joy
Jan 23 at 18:37
$begingroup$
@JohnJoy I worked it out your way, and I get two possible solutions for an angle, with one of them being correct after disproving the other. However, I think that this amount of calculation is much beyond the level of this problem, and I suspect there's a simpler solution. Nevertheless, this works very well — thank you.
$endgroup$
– daedsidog
Jan 23 at 18:58