One Angle Twice the Other

From the sine theorem, we have $\frac{x}{\sin 2\alpha} = \frac{13}{\sin \alpha} = \frac{31}{\sin(180^\circ - 3\alpha)} = \frac{31}{\sin\ 3\alpha}$

$4 \cos^2 \alpha - 1 = \frac{\sin 3\alpha}{\sin \alpha} = \frac{31}{13}$

$2\cos \alpha = \frac{\sin 2\alpha}{\sin \alpha} = \frac{x}{13}$

Squaring the second one and subtracting them, we get $x^2 = 572$

Let $\angle ABC=\theta$ . $\angle BAC=2\theta$ , $\angle ABC=180^\circ-3\theta$ . Using $\sin(180^\circ-x)=\sin x$ and by sine rule, $\frac{31}{\sin3\theta}=\frac{13}{\sin\theta}$ . Since $\sin3\theta=(\sin\theta)(3-4\sin^2\theta)$ , $\sin^2\theta=\frac{2}{13}$ . $\cos2\theta=1-2\sin^2\theta=1-2(\frac{2}{13})=\frac{9}{13}$ . By Cosine rule, $BC^2=31^2+13^2-2(31)(13)(\cos2\theta)=1130-806(\frac{9}{13})=572$ .

Consider the point $D$ on $BC$ such that $AC$ bisects $\angle BAC$ . The angle bisector theorem gives us $\frac{|DB|}{|DC|} = \frac{|AB|}{|AC|} = \frac{31}{13}$ so let $|DB|=31k, |DC|=13k$ for some $k \in \mathbb{Z}^+$ .

Note that $\angle DAC = \frac{1}{2}\angle BAC = \angle ABC = \angle DBC$ so $\Delta DAB$ is isosceles where $|DA|=|DB|=31k$ . Using Stewart's Theorem with cevian $DA$ then gives us, $|DB||AC|^2 + |DC||AB|^2 = |BC|(|DA|^2+|DB||DC|$ $\Rightarrow (31k)(13^2)+(13k)(31^2) = (44k)((31k)^2+(31k)(13k))$ $\Rightarrow 17732k = 60016k^3$

As $k \neq 0$ , we get $k^2 = \frac{13}{44}$ . Therefore, $|BC|^2 = (44k)^2 = 44*13 = 572$

let $\angle ABC$ = X,

therefore $\angle BAC$ = 2X

$\cos X = \frac{ 31^2 + BC^2 - 13^2 }{ 2*31*BC }$ ......(1)

and $\cos 2X = \frac{ 31^2 + 13^2 - BC^2 }{ 2*31*13 }$ .......(2)

and as $\cos 2X = 2*(cos X)^2 - 1$ .......(3)

therefore by equating (3) & (2) :-

$\sqrt{2}cos X = \sqrt{ 1+ \frac{31^2 + 13^2 - BC^2}{2*31*13}}$ ......(4)

and by equating (1) & (4) :-

we get , $BC^2$ = 572 or 324

but $BC^2$ cannot be equal to 324 as then BC = 18 and thus no triangle is possible as $BC + AC = 18 +13 = 31$ that is equal to the 3rd side AB.

therefore, $BC^2 = 572$

Let $\angle ABC= x, \angle BAC=2x, \angle ACB=180-3x$ . Using law of sines on sides AC and AB, $\frac{sinx}{13} =\frac{sin(180-3x)}{31}$ . We know that $sin(180-3x)=sin3x=4sinx-3sin^3x$ . Simplifying the above equation, we have $sin^2 x=\frac{2}{13}$ . Now, using law of cosines, $BC^2 =13^2+31^2-2(13)(31)cos2x$ , where $cos2x=1-2sin^2x=\frac{9}{13}$ . Hence, $BC^2=572$ .

Let the angle bisector of $\angle A$ hit $\overline{BC}$ at $D$ . Note that $\angle ABC=\angle DAB\implies \overline{AD}=\overline{BD}$ . Let $\overline{AD}=x$ and $\overline{CD}=y$ . Note that $\Delta ACD\sim\Delta ABC$ because of AA similarity. Therefore, $\dfrac{x}{13}=\dfrac{31}{x+y}$ and $\dfrac{y}{13}=\dfrac{13}{x+y}$ . Clearing denominators we get $x^2+xy=403$ and $y^2+xy=169$ . Add these two equations together, and we get $x^2+2xy+y^2=572$ . Recall that $x^2+2xy+y^2=(x+y)^2$ , so $(x+y)^2=572$ . Since $x+y=\overline{BD}+\overline{CD}=\overline{BC}$ , we are conveniently done. the answer is therefore $\boxed{572}$ .

Let $\angle ABC = \theta$ . Then $\angle ACB = 180 - 3 \theta$ . Now by Law of Sines we have: $\frac{31}{\sin 3 \theta} = \frac{13}{\sin \theta}$ Now via $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$ we can solve for $\sin \theta$ and get $\sin \theta = \sqrt{\frac{2}{13}}$ . This implies $\cos 2 \theta = 1 - \frac{4}{13} = \frac{9}{13}$ . Now by Law of Cosines we get $BC^2 = 13^2 + 31^2 - 2 \cdot 13 \cdot 31 \cdot \frac{4}{13} = 572$ as desired.

Draw out triangle ABC. Angles ABC and BAC have to be acute or the triangle's angles and side lengths won't match up (biggest angle gets longest side etc...). Label side BC as x

Next, use the Law of Sines. sin(2a) / x = sin(a) / 13 sin(2a) = 2sin(a)cos(a) x = 26cos(a)

Next, use the Law of Cosines. x^2 = 31^2 + 13^2 - 2 13 31*cos(2a) 676cos^2(a) = 1130 - 806 * (2cos^2(a) - 1) 2288cos^2(a) = 1936 cos^2(a) = 1936/2288

BC^2 = x^2 = 26^2 * cos^2(a) = 572

Let $D$ be the point on $BC$ such that $AD$ bisects angle $BAC$ . Then $ADB$ is isosceles so $BD = AD$ , and triangle $ADC$ is similar to $BAC$ . The ratio of similarity of $BAC$ to $ADC$ is equal to $BC/13$ as well as $31/AD$ , so $BC \cdot AD = 13 \cdot 31$ .

If we repeat the bisection on angle $ADC$ by placing point $E$ on $AC$ we get another similar triangle $DEC$ whose sides are parallel to $BAC$ . Via repetition, the ratio of similarity of $BAC$ to $DEC$ must be $(BAC:ADC)^2 = (BC/13)^2$ , but it's also equal to $BC/CD$ , so $BC \cdot CD = 13^2$ . Now we just add: $BC^2 = BC \cdot (BD + DC) = BC \cdot (AD + CD) = 13 \cdot 44 = 572.$

In order to find the value of BC^2 , we must first know the value of \cos \angle BAC since as what cosine rule states, the square of the length of the side of a triangle is equal to the sum of the squares of the two other side minus twice the product of the two other side and the cosine value of its opposite angle.

Let x denote the value of \angle ABC which then make the value of \angle BAC equal to 2x. Solving for the value of \angle ACB we will have 180 - 3x.

  By Sine Rule we will have  \frac {13}{\sin x} = \frac {31}{\sin (180 - 3x)}

By sum and difference formula for sine, \sin (180 - 3x) will be (\sin 180) (\cos 3x) - (\sin 3x) (\cos 180) . Simplifying this expression will lead to a value of \sin 3x .

So the equation now becomes \frac {13}{\sin x} = \frac {31}{\sin 3x} which can be shown as \frac {31}{13} = \frac {\sin 3x}{\sin x} \frac {31}{13} = \frac {(\sin 2x)(\cos x) + (\sin x)(\cos 2x)}{\sin x} \frac {31}{13} = \frac {(2\sin x (\cos x)^2 + (\sinx)[(\cos x)^2 - (\sin x)^2]}{\sin x}

Simplifying the right hand side, we will have:

\frac {31}{13} = 2(\cos x)^2 + {(\cos x)^2 - [1 - (\cos x)^2]} \frac {31}{13} = 4(\cos x)^2 - 1 [\frac {31}{13} + 1]/4 = (\cos x)^2

We need to have the cosine value of the opposite angle of side BC, which is \angle BAC that we denoted as 2x.

since \cos 2x = 2\cos x)^2 - 1, we need to multiply our equation by two and subtract 1 at both sides to get the value of \cos 2x.

Our equation will now be {[\frac {31}{13} + 1]/2} - 1 = \cos 2x

Simplifying, we will have \frac {9}{13} = \cos 2x

By cosine rule, BC^2 = (AB)^2 + (AC)^2 - 2(AB)(AC)(\cos \angle BAC) BC^2 = (AB)^2 + (AC)^2 - 2(AB)(AC)(\cos 2x) BC^2 = (31)^2 + (13)^2 - 2(13)(31)(\frac {9}{13}) BC^2 = 961 + 169 - 558 BC^2 = 572

For ease of writing, let $x = BC$ .

Construct the internal angle bisector of $\angle BAC$ and let $D$ be the point of intersection of the angle bisector and $\overline{BC}$ . By the angle bisector theorem, $\frac{AC}{AB} = \frac{DC}{DB} = \frac{13}{31}$ , which means that $DC = \frac{13}{44}x$ and $DB = \frac{31}{44}x$ . Because $\textrm{m} \angle BAC = 2\textrm{m} \angle ABC$ , we have that $\textrm{m} \angle BAD = \textrm{m} \angle CAD =$ $\textrm{m} \angle BAC$ , and so $\bigtriangleup ABD$ is isosceles and $AD = BD = \frac{31}{44}x$ . The previous line also gives us that $\bigtriangleup BAC \sim \bigtriangleup ADC$ by Angle-Angle similarity.

We are ready to complete the problem: by similar triangles, $\frac{AC}{DC} = \frac{BC}{AC} \to AC^2 = BC \cdot DC \to$ $169 = x \cdot \frac{13}{44}x \to x^2 = 44 \cdot 13 = \fbox{572}$ .

Let x be the measure of angle ABC, so that 2x is the measure of angle BAC and 180-3x is the measure of angle ACB. By the Law of Sines, AB/sinC=AC/sinB, or 31/(sin(180-3x))=31/sin(3x)=13/sin(x). Since sin(3x)=3sin(x)-4[sin(x)]^3, we substitute and multiply both sides by sin(x) to get

13=31/(3-4[sin(x)]^2), or 39-52[sin(x)]^2=31, or [sin(x)]^2=2/13.

By the Law of Cosines, BC^2=AB^2+AC^2-2(AB)(AC)cosA=31^2+13^2-2(31)(13)cos(2x)=31^2+13^2-2(31)(13)[1-2[sin(x)]^2]=961+169-2 31 9=572, as desired.

By using the trigonometric area formula for triangles $Area = \frac {ab\sin C}{2}$ . If we set $BC = x$ and $\angle ABC = \theta$ . Then using the the area formula on both $\angle ABC$ and $\angle BAC$ we have $\frac {31x\sin \theta}{2} = \frac{31 \times 13\sin 2\theta}{2}$ . Which simplifies down to (using the double angle formula for sine) $\frac {x}{26} = \cos \theta$ . Now using the law of cosines on $\angle ABC$ we have

$31^2 + x^2 - 2\times 31x\cos \theta = 13^2$

Simplifying and plugging in $\frac {x}{26} = \cos \theta$ we get

$792 = \frac {18x^2}{13}$

Therefore $x^2 = BC^2 = 572$

The answer is $572.$

By using $Sine Rule,$

$L_{ABC} = \frac{1}{2} \times AB \times AC \times \sin ∠BAC = \frac{1}{2} \times AB \times BC \times \sin ∠ABC$

$\frac{1}{2} \times 31 \times 13 \times \sin 2∠ABC = \frac{1}{2} \times 31 \times BC \times \sin ∠ABC$

$\cos ∠ABC = \frac{BC}{26}$

Again, by using $Cosine Rule.$

$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos ∠ABC$

$13^2 = 31^2 + BC^2 - 2\times 31 \times BC \times \frac{BC}{26}$

$BC^2 = 572$

Let angle ABC be x Using the sine rule, 13/sinx = 31/sin(pi - 3x) This gives us x to be root 11/13. From here, we will use another sine rule BC/sin(2root11/13) = 13/sin(root11/13) BC = 26 root 11/13 BC^2 = 572

consider angle ABC is a, and angle BAC is 2a. now use the cosine formula to calculate cos(a) and cos(2a) In a triangle ABC, cos(A)=(b^2 + c^2 - a^2)/2bc and equate the cos(2a) as 2cos^(a) - 1

now this is an algebraic problem, we have to solve a bi-quadratic equation. notice that here power of side BC is even always (0,2,or 4). consider BC^2 as x, and solve a quadratic equation, you will get your answer.

We naming the angle $\angle CAB = 2\alpha$ and the $\angle CBA = \alpha$ . Build the line $CM$ , where the $\angle CMB = 2\alpha$ , then $CM = 13$ , as $\angle ABC + \angle BCM = \angle AMC$ then $\angle BCM = \alpha$ . So $\triangle MCB$ is isosceles, $MB = 13$ and $AM = 18$ . We build the perpendicular $CP$ , with foot in $AB$ . Using the theorem of pitagoras in $\triangle CPM$ we find that $CP = \sqrt{88}$ . Using theorem of pitagoras in $\triangle CPB$ we find that $CB^2 = 572$ .

BC^2=13^2+31^2-2 * 13 * 31*cosX.
BC/sinX=13/sin(X/2).
Solve it you will get BC^2=572

$BC^{2}$ = $AB^{2}$ + $AC^{2}$ - 2AB.AC.cos(2x) (*) with x=∠ABC

Use the formula : $\frac{a-b}{a+b}$ = $\frac{tan( \frac{A-B}{2} )}{tan( \frac{A+B}{2} )}$

we have $\frac{31-13}{31+13}$ = $\frac{tan( \frac{180-3x-x}{2} )}{tan( \frac{180-3x+x}{2} )}$ <=> $\frac{9}{22}$ = $\frac{tan(x)}{tan(2x)}$ => $tan(x)^{2}$ = $\frac{2}{11}$ => $cos(x)^{2}$ = $\frac{11}{13}$ => cos(2x)= $frac{9}{13}$ from (*) => $BC^{2}$ = 572

Let BC=x

Let angle BAC=2a

Let angle ABC=a

From trig identities, we know that cos2a=2cos^2x-1

From the cosine law,

x^2=1130-806cos(2a)

x^2=1130-806(2cos^2x-1)

From the sine law,

sin(2a)/x = sina/13

2sinacosa/x=sina/13

x=26cosa

x^2=676cos^2x

Subbing this into the other equation gives,

676cos^2x=1130-806(2cos^2x-1)

Solving for cosx, we get cosx=root(11/13)

Subbing this back into the original equation, we get,

x^2=1130-806(2(11/13)-1)

x^2=572

Therefore BC^2=572

In this solution, $∠ABC$ will be called $α$ . Therefore, $∠BAC = 2α$ . We will also refer to $BC$ as $x$ .

First of all, we will use the law of sines:

$\frac{x}{sin(2α)} = \frac{13}{sin(α)}$

Knowing that $sin(2α) = 2sin(α)cos(α)$ , we get to:

$\frac{x}{13} = 2cos(α)$

Now, we will use the law of cosines:

$x^{2} = 31^{2} + 13^{2} - 2 \times 31 \times 13 \times cos(2α)$

Knowing that $cos(2α) = 2cos^{2}(α) - 1$ , we have:

$x^{2} = 1936 - 1612cos^{2}(α)$

Plugging in:

$x^{2} = 1936 - \frac{403x^{2}}{169}$

$x^{2} = 572$

First, draw the angle bisector of $\angle A$ and call the intersection point on line $BC$ $D$ .

We know that $\angle DAB \cong \angle DBA$ because of the angle bisector, so therefore $AD=BD$ . Call angle $\angle DAB=x$ . From there, we can tell that $\angle ADC=2x$ .

Analyzing the figure, we can see that $\angle CAD=x=\angle ABD$ . We also know that $\angle BAD=\angle ADC$ . This means that $\bigtriangleup ADC~ \bigtriangleup BAC$ . This implies that $\frac{CB}{13}=\frac{13}{CD} \Longrightarrow CB*CD=169$ . By the angle bisector theorem, $\frac{13}{CD}=\frac{31}{BD} \Longrightarrow 13BD=31CD$ .

Solving for, we get BD= $\frac{31CD}{13}$ . Now substitute in BD into the first equation:

$CD(\frac{31CD}{13}+\frac{13CD}{13}=169)$ . Therefore, $CD=\frac{13\sqrt{143}}{22}$ . Solving for BC, we get $\frac{31\sqrt{143}}{22}$

Adding them and squaring, we get $\boxed{572}$ .

First, let's use triangle square formula: $\frac{1}{2} \times 31 \times BC \times \sin \alpha = \frac{1}{2} \times 31 \times 13 \times \sin 2\alpha$

$\sin 2\alpha$ = $2 \times \sin \alpha \times \cos \alpha$

So:

$\cos \alpha = \frac{BC}{26}$

Next let's use Law of cosines :

$BC^{2} = 31^{2} + 13^{2} - 2 \times 13 \times 31 \times \cos 2\alpha$

Using previous conclusion and $\cos 2\alpha = 2\cos^{2}\alpha-1$ , we can achieve $BC^{2} = 572$

Because angle BAC is twice angle ABC, if you draw the angle bisector of angle BAC and let it hit BC at D, you get angle DAB is equal to angle DBA, so BD=DA. Using the angle bisector theorem, AB/BD=AC/CD, so we can use multipliers of 31 AB and AC for BD and DC, so we can let BD=31x and let DC=13x, because AB=31 and AC=13. We have the lengths of the sides and the length of a Cevian, so we can use Stewart's Theorem to find the value of x. 31 44 13 x^3+31 31 44 x^3=31 31 13 z+13 13 31 x. Divide by x to get 31 44 13 x^2+31 31 44 x^2=31 31 13+13 13 31. Factor out 31 44 x^2 from the left side and 31 13 from the right side to get 31 44 x^2(13+31)=31 13(31+13). 13+31=31+13 by the commutative property of addition, and they are both equal to 44, so we can divide 44 from both sides to get 31 44 x^2=31 13. Divide 31 from both sides to get 44 x^2=13, so x^2=13/44. Note that since BD=31x and DC=13x, BD+DC=31x+13x=(31+13)x=44x. BC^2=(44x)^2=44 44 x^2. From the Stewart's equations, we have x^2=13/44, so BC^2=44 44 13/44, and 44/44=1, so we have 44 13 1=44*13=572.

The third angle, opposite AB is (180 - 3theta) Using law of sines we can make 13/sin(theta) = sin(180 - 3theta)/ 31. Sin(180 - x) = sinx so sin(180 - 3theta) = sin(3theta). sin(3theta) = 3sin(theta) - 4sin^3(theta). This can be proven by using sin(2theta + theta). Rearranging law of sines from earlier, it can be concluded 31sin(theta) = 13(3sin(theta) - 4sin^3(theta)) ---> 52sin^3(theta) -8sin(theta) = 0, 4sin(theta)(13sin^2(theta)-2) = 0, 4sin(theta) can't be 0 because that would make theta equal 0 degrees which isn't possible in a triangle. So, 13sin^2(theta) - 2 = 0. sin(theta) = sqrt(2)/ sqrt(13). cos(theta) = sqrt(11)/sqrt(13) which can be found from the identity sin^2(theta) + cos^2(theta) = 1. Then we can use law of sines again, BC/sin(2theta) = 13/sin(theta) ---> BC = 13sin(2theta)/sin(theta), 13(2sin(theta)cos(theta))/sin(theta), 13(2(sqrt(22)/13))/sqrt(2)/sqrt13, BC = 2(sqrt(11)sqrt(13)), BC^2 = 4 * 11 * 13 = 572.

Draw the angle bisector of $\angle BAC$ and let $D$ intersection with $BC$ . Call $BD=x$ and $CD=BC-x$ Applying Stewart's theorem modified by the angle bisector, we get $AD^{2}+x \times (BC-x) =31 \times 13$ , which simplifies to $x \times BC=403 (1)$ . Also, from the angle bisector theorem, we get $31/x=13/(BC-x) (2)$ . Solving this equation for x and plugging into $(1)$ , we get $BC=572$

We're going to call $\angle ABC$ , $\theta$ for the sake of convenience.

So, $\angle ABC=\theta$

$\angle BAC= 2\theta$

And $\angle ACB= 180^{\circ} - 3\theta$ .

We want $BC^2$ .

From the cosine law,

$BC^2={AB}^2+{AC}^2-2AB\times AC \times \cos \angle BAC$ .

$\Rightarrow BC^2={31}^2+{13}^2-2\times 31\times 13 \times \cos 2\theta$

$\Rightarrow BC^2= 961 +169 -806(\cos^2 \theta-\sin^2 \theta)$

$\Rightarrow BC^2= 1130 -806(1- 2\sin^2 \theta)$ $\cdots (1)$ .

We are going to keep it this way.

Now let's apply the sine law on $\triangle ABC$ . It tells us:

$\frac{AB}{\sin (180^{\circ} - 3\theta)}=\frac{AC}{\sin \theta}$ .

$\Rightarrow \frac{31}{\sin 3\theta}=\frac{13}{\sin \theta}$ .

We're going to substitute $\sin 3\theta$ with $(3\sin\theta-4\sin^3\theta)$ .

So that gives us:

$\frac{31}{3\sin\theta-4\sin^3\theta}=\frac{13}{\sin \theta}$ .

A little bit of re-arranging gives us:

$52\sin^3\theta - 8\sin\theta=0$

$4\sin\theta(13\sin^2\theta-2)=0$

implying $\sin\theta=0$ or $13\sin^2\theta-2=0$ . So we are left with two cases.

Case $1$ : [ $\sin\theta=0$ ]

Go back to $(1)$ which tells us

$BC^2= 1130 -806(1- 2\sin^2 \theta)$ .

Since $\sin \theta =0$ , $\sin^2 \theta =0$ as well.

Plugging that in gives us

$BC^2=324$ .

That may look like a good solution, but notice that $BC$ is now equal to $18$ and $AB=BC+AC$ which makes $\triangle ABC$ degenerate.

Case $2$ : [ $13\sin^2\theta-2=0$ ]

In this case $\sin^2 \theta=\frac{2}{13}$ .

Plugging that in $(1)$ gives us:

$BC^2=572$ . This doesn't have issue case $1$ had.

So, $BC^2=\boxed{572}$ .

Law of sines: 31/sin3x = 13/sinx -> sin3x = sinx(3-4sin²x) So: sinx = sqrt(2/13) -> cos2x = 1 - 2sin²x = 9/13. Law of cosines: BC² = AB² + AC² - 2(AB)(AC)cos2x .:. BC² = 572