Squaring a Leg

Draw $AD$ where $D$ is on $BC$ and $AD$ bisects $\angle BAC$ .

Thus $\angle BAD=\angle ABC$ i.e. $\Delta ABD$ is isosceles.

$\angle ADC=\angle BAD+\angle ABC=\angle BAC$ , hence $\Delta ABC\sim\Delta DAC$ .

This means that $\frac{AB}{AC}=\frac{DA}{DC}=\frac{DB}{DC}$ . Let $BC=a$ hence $DB=\frac{25}{36}a$ and $DC=\frac{11}{36}a$ .

Since $\frac{BC}{AC}=\frac{AC}{DC}$ , we form this equation $\frac{a}{11}=\frac{11}{\frac{11}{36}a}$ .

Manipulate to get $a^2=396$ , which is the answer.

Name angles and sides of the triangle: $\begin{array}{l l l l} &A &= &\angle BAC \\ &B &= &\angle ABC \\ &C &= &\angle BCA \\ &a &= &BC \\ &b &= &AC \\ &c &= &AB \end{array}$

This implies the following:

$\begin{array}{l l l l} &A &= &2B \\ &C &= &180^ \circ-A-B \\ & &= &180^ \circ-3B \\ &b &= &11 \\ &c &= &25 \end{array}$

Law of Sines: $\begin{array}{l l l l l l} &\frac {a}{\sin A} &= &\frac {b}{\sin B} &= &\frac {c}{\sin C} \\ \Rightarrow & & &\frac {11}{\sin B} &= &\frac {25}{\sin (180^ \circ-3B)} \\ \end{array}$

Difference Formulas: $\begin{array}{l l l l} &\sin (\alpha-\beta) &= &\sin \alpha \cos \beta - \cos \alpha \sin \beta \\ \Rightarrow &\sin (180^ \circ - \theta) &= &\sin \theta \\ \Rightarrow &\sin (180^ \circ - 3B) &= &\sin 3B \end{array}$

Triple Angle Formulas: $\begin{array}{l l l l} &\sin 3\theta &= &3\sin \theta - 4 \sin^3 \theta \\ \Rightarrow &\sin 3B &= &3\sin B - 4\sin^3 B \end{array}$

Continuing the solution, $\begin{array}{l l l l} \Rightarrow &\frac {11}{\sin B} &= &\frac {25}{\sin 3B} \\ & &= &\frac {25}{3\sin B - 4 {\sin}^3 B } \\ \Rightarrow &\sin^2 B &= &\frac {2}{11} \end{array}$

Law of Cosines: $\begin{array}{l l l l} &a^2 &= &b^2 + c^2 - 2 \cdot b \cdot c \cdot \cos A \\ & &= &11^2 + 25^2 - 2 \cdot 11 \cdot 25 \cdot \cos 2B \end{array}$

Double Angle Formulas: $\begin{array}{l l l l} &\cos 2\theta &= &1 - 2 \sin^2 \theta \\ \Rightarrow &\cos 2B &= &1 - 2 \sin^2 B \\ & &= &1 - 2 \cdot \frac{2}{11} \\ & &= &\frac{7}{11} \end{array}$

Continuing the solution,
$\begin{array}{l l l l} &a^2 &= &11^2 + 25^2 - 2 \cdot 11 \cdot 25 \cdot \frac{7}{11} \\ & &= &396 \end{array}$

We draw the angle bisector of $\angle{BAC}$ . Suppose that it hits $\overline{BC}$ at $D$ . Now $\angle{DAB} = \angle{DBA}$ , so $\triangle{DAB}$ is isosceles. Let $DA=DB=X$ . By the angle bisector theorem, $DC = \frac{11}{25}x$ . Now, by Stewart's theorem for angle bisectors, we have $x^2=\frac{11\cdot25}{36^2}(36^2-(\frac{36}{11}x)^2) \Rightarrow {BC}^2 = (\frac{36}{25}x)^2 = \frac{36^2 \cdot 275 \cdot 625}{25^2 \cdot 900} = 396$ .

Let $D$ be a point on line $BC$ such that $AD$ bisects $∠BAC$ . Note that $∠DAB=∠ABD$ so triangle $ABD$ is isosceles, where $AD=DB$ .

So now we have $∠CAD=∠CBA$ and $∠CDA=∠CAB$ which implies that triangles $ADC$ and $BAC$ are similar. So we have;

$\frac{AC}{BC}=\frac{DA}{AB}=\frac{DC}{AC}$

or

$\frac{11}{DC+BD}=\frac{BD}{25}=\frac{DC}{11}$

So we can conclude that $BD=\frac{25}{11}DC$

So, $\frac{11}{DC+\frac{25}{11}DC}=\frac{DC}{11}\Rightarrow DC^2=\frac{11^2}{(\frac{36}{11})}$

Now $BC^2=(DC+DB)^2$ $=(DC+\frac{25}{11}DC)^2$ $=(\frac{36}{11}DC)^2$ $=(\frac{36}{11})^2\frac{11^2}{(\frac{36}{11})}$ $=396$

Let angle BAC be 2x and angle ABC be x, then angle ACB will be (180-3x). By cosine rule, BC^2=25^2+11^2-2(25)(11)cos2x. By sine rule, 25/sin(180-3x)=11/sinx. Since sin(180-3x)=sin3x=(sinx)(3-4sin^2x), by simplifying the sine rule equation, we get sin^2 x=2/11. Since cos2x=1-2 sin^2 x, by substituting sin^2 x=2/11 into cos2x, we get cos2x=7/11. Hence, by substituting cos2x=7/11 into the cosine rule equation, we can get BC^2=396.

We have, in triangle ABC, $\angle BAC = 2 \times \angle ABC$ Let $\angle ABC = x$ , so $\angle BAC = 2x$ . By angle sum property, $\angle BCA = \pi - 3x$ . Again let BC = $y$ .Now by applying SINE RULE, area of triangle ABC = $\frac {1}{2} AC \times BC \times sinC$ = $\frac {1}{2} \times 11 \times y \times sin( \pi - 3x$ ) = $\frac {1}{2} \times 11 \times y \times sin( 3x$ ) because $sin( \pi - 3x$ ) = $sin(3x)$ . Now, by applying SINE RULE once again, area of triangle ABC = $\frac {1}{2} AB \times BC \times sinB$ = $\frac {1}{2} 25 \times y \times sinx$ . Since area of triangle is equal if we calculate it in any way. So, $\frac {1}{2} 11 \times y \times sin3x$ = $\frac {1}{2} 25 \times y \times sinx$ . \Rightarrow \(11 \times sin3x = 25 \times sinx Putting $sin3x = 3 sinx - 4 (sinx)^3$ in the above equation we get $11 ( 3 sinx - 4 (sinx)^3) = 25 sinx$ $\Rightarrow 44(sinx)^3 = 8 sinx$ . So this means either $44 (sinx)^2 = 8$ OR $sinx = 0$ . But $sinx$ can't be 0 as x can't be 0, $\pi$ etc. for a triangle. Hence, $44 (sinx)^2 = 8$ $\Rightarrow (sinx)^2 = \frac {8}{44}$ ....................... ( i ) Now once again taking SINE RULE, area of triangle ABC = $\frac {1}{2} AC \times AB \times sinA$ = $\frac {1}{2} 11 \times 25 \times sin2x$ . Now equating it with area of triangle earlier derived we get $\frac {1}{2} 25 \times y \times sinx$ = $\frac {1}{2} 11 \times 25 \times sin2x$ $\Rightarrow y sinx = 11 sin2x$ $\Rightarrow y sinx = 11 \times 2 \times sinx \times cosx$ $\Rightarrow y = 22 cosx = 22 \sqrt{1 - (sinx)^2} = 22 \sqrt{1 - \frac {8}{44}} = 22 \sqrt{\frac {36}{44}}$ $\Rightarrow y = 6 \sqrt{11}$ . Here $cosx$ can't be negative as it will make $y$ negative. so, $y = 6 \sqrt{11}$ . Hence, $y^2 = (BC)^2 = 396$ . So 396 is the required answer.

Firstly, let us label Angle ABC, x. Therefore, Angle BAC = 2x.

Drawing the triangle, and using the Sin Rule, we immediately see that,

\frac {11}{sin x} = \frac {BC}{sin 2x}

\Rightarrow BC = \frac {11 sin 2x} {sin x}

But, as sin 2x = 2 sin x cos x,

BC = 22 cos x.

Now, using the cos rule,

BC^2 = 25^2 + 11^2- 2(275)(cos 2x)

But, cos 2x = 2cos^2 x - 1, and, by the relation given above, cos x = BC/ 22.

So, BC^2 = 625+121 - 2(275)(\frac {2*BC^2}{484} - 1)

Simplifying, we have,

\frac {36 BC^2} {11} = 1296

Which leaves us with,

BC^2 = 396

Let \angle ABC = x ^ \circ and \angle BAC = 2x ^ \circ. Hence, \angle BCA = 180-3x ^ \circ

By Sine Rule, we have: \frac {BC}{sin A} = \frac {AB}{sin C} = \frac {AC}{sin B}

=> \frac {BC}{sin 2x} = \frac {25}{sin 3x} = \frac {11}{sin180-3x}

Solving the equalities separately, and after simplifying we get: cos 2x=25/11 - BC^2/242 ....(i)

Using cosine rule, we get BC^2=AC^2 + AB^2 + 2(AB)(AC)(cosA)

Using (i) to substitute for cosA, we can solve for BC^2 to get BC^2 as 396

Let the angle bisector of $\angle CAB$ hit $BC$ at $D$ .

Now let $BC=x$ .

By the angle bisector theorem,

$\frac{CD}{DB}=\frac{CA}{AB}$

So

$CD=CB\cdot \frac {CA}{CA+AB}=\frac {11}{36}x$

$DB=CB\cdot \frac {AB}{CA+AB}=\frac {25}{36}x$

By Stewart's Theorem,

$AD^2=AC\cdot AB - DC \cdot DB$

Thus

$AD^2=275 - \frac{275}{36^2}x^2$

Also since $\angle DAB= \frac12 \angle CAB = \angle DBA$ ,

we have $DB=DA$ .

Thus

$\begin{aligned} 275 - \frac{275}{36^2}x^2&=AD^2=DB^2=(\frac {25}{36}x)^2 \\ 275-\frac{275}{36^2}x^2&=\frac{25^2}{36^2}x^2\\ 275&=\frac{275+625}{36^2}x^2\\ 275&=\frac{900}{36^2}x^2\\ x^2&=275\cdot \frac{1296}{900}=396. \end{aligned}$

Let the angle bisector of $A$ intersect $BC$ at $D$ . Let $\angle CBA = \angle BAD = \angle DAC = \alpha$ , and $AD=BD=x$ . Clearly, $\angle ADC = 2\alpha$ .

Now, by angle bisector theorem, we have $\frac{25}{x} = \frac{11}{DC} \Rightarrow DC = \frac{11}{25}x$ .

Next, applying sine rule on $\triangle ABC$ we have $\frac{\frac{36}{25}x}{\sin 2 \alpha} = \frac{11}{\sin \alpha}$ . And applying sine rule on $\triangle ABD$ , we have $\frac{x}{\sin \alpha}$ = \frac{25}{\sin (180^\circ - 2 \alpha) $= \frac{25}{\sin 2 \alpha}$ . Multiplying these two equations and getting rid of the common denominators ( $\sin\alpha \sin 2 \alpha$ ), we have that $\frac{36}{25}x^2 = 275$ .

Now, the questions requires we find $BC^2 = (\frac{36}{25}x)^2 = (\frac{36}{25})(\frac{36}{25}x^2) = (\frac{36}{25})(275)=396$ .

Let angle ABC be x ,then angle BAC is 2 x. By sine rule , (sin x)/11 = (sin 2 x)/BC by cosine rule , cos x = (25^2 + BC^2 - 11^2)/2(25)(BC) By solving the two equations answer comes out to 396

Let $\angle{BAC} = 2x$ , $\angle{ABC} = x$ .

By sine rule, $\frac{11}{sinx} = \frac{BC}{sin2x}$

=> $\frac{11}{sinx} = \frac{BC}{2sinxcosx}$

=> $cos x$ = $\frac{BC}{22}$

By cosine rule, $BC^2 = 11^2 + 25^2 - 2(11)(25)(cos2x)$

Consider $cos2x$ = $2cos^2x - 1$ . By substituting $cosx$ previously obtained, we have :

$cos2x$ = $\frac{1}{242} BC^2 - 1$

Hence, $BC^2 = 1296 - \frac{25}{11}BC^2$

$\frac{36}{11}BC^2 = 1296$

$BC^2 = \boxed{396}$

just apply sine rule........

Let $\angle{ABC}=\theta$ and let $BC=x$ . Then $\angle{BAC}=2\theta$ . We want the value of $x^2$ ,

Firstly, we use the Law of Sines in $\triangle{ABC}$ to find

$\frac{11}{\sin\theta}=\frac{x}{\sin{2\theta}}=\frac{x}{2\sin\theta\cos\theta} \Rightarrow 22\sin\theta\cos\theta=x\sin\theta \Rightarrow \cos\theta=\frac{x}{22}\cdots(1)$

Now we use the Law of Cosines in $\triangle{ABC}$ to find

$11^2=25^2+x^2-2(25)x\cos\theta \Rightarrow \cos\theta=\frac{(25-11)(25+11)+x^2}{2(25)x}=\frac{x^2+(14)(36)}{50x}\cdots(2)$

Now we let $(1)=(2)$ to obtain

$\cos\theta=\frac{x}{22}=\frac{x^2+(14)(36)}{50x} \Rightarrow 50x^2=22x^2+22(14)(36) \Rightarrow BC^2=x^2=\frac{(22)(14)(36)}{2(14)}=20^2-2^2=\boxed{396}$

Let $BC = x$ and let $\angle ABC = \beta$ . Then, from the Law of Cosines and the Law of Sines, respectively, we get: $\begin{aligned} x^2 &=& 625 + 121 - 50 \cdot 11 \cdot \cos (2\beta), \\ \dfrac {\sin(2\beta)}{x} &=& \dfrac {\sin(\beta)}{11}. \end{aligned}$ Simplifying the second equation, we get $22 \cos (\beta) = x.$ From the first equation, we get $x^2 = 746 - 550 \cdot \left( 2 \cos^2 (\beta) - 1 \right) = 484 \cos^2 (\beta) \implies \cos^2 (\beta) = \dfrac {9}{11}.$ Thus, we have $x^2 = 484 \cos^2 (\beta) = 484 \cdot \dfrac {9}{11} = \boxed {396}$ .

BC^2 = AC(AC+AB)=11(25+11)=11×36=396

Let angle ABC =x, then angle BAC = 2x and angle ACB = 180 - 3x.

Using sine law, we have

sin 2x / BC = sin x / 11 =sin 3x / 25

Now, sin x / 11 = sin 3x / 25 implies sin 3x / sin x = 25/11 implies

3 - 4 sin^2 x = 25 / 11 implies 4 cos^2 x - 1 = 25/11 implies 4 cos^2 x = 36 / 11

and sin 2x / BC =sinx /11 implies BC = 11 sin 2x / sin x = 11 * 2 cos x

Therefore BC^2 = 121 * 4 cos^2 x = 121 * 36 /11 = 396.

Hence, BC^2 = 396 is the required solution.

• We know The Law of Cosines: $AB^{2} = BC^{2} + AC^{2} - 2.AC.BC.cos\angle{ACB}$
• $\angle{ABC} = \theta => \angle{BAC} = 2\theta$

Then, we have 2 equations:

• $BC^{2} = AC^{2} + AB^{2} - 2AB.AC.cos\angle{2\theta}$ (named as equation 1)
• $AC^{2} = AB^{2} + BC^{2} - 2AB.BC.cos\angle{\theta}$

<=>

• $BC^{2} +550(2cos^{2}\theta -1) = 746$
• $BC^{2} - 50BC.cos\theta = -504$

<=>

• $BC^{2} + 1100cos^{2}\theta = 1296$
• $\frac{-18BC^{2}}{7} + \frac{900BC.cos\theta}{7} = 1296$

=>

$BC^{2} + 1100cos^{2}\theta = \frac{-18BC^{2}}{7} + \frac{900BC.cos\theta}{7}$

<=>

$\frac{7700cos^{2}\theta}{BC^{2}} - \frac{900cos\theta}{BC^{2}} + 25 = 0$

=>

• $\frac{cos\theta}{BC} = \frac{1}{14}$
• $\frac{cos\theta}{BC} = \frac{1}{22}$

Compare with equation 1, we have:

• $BC_{1} = 14$ (wrong because it doesn't satisfied Triangle inequality)
• $BC_{2} = 6\sqrt{11}$ (satisfied Triangle inequality)

=> $\boxed{BC^{2} = 396}$

$\angle\ C=180 - (B + 2B)=180 - 3B.\ \ \ Sin3B= - 4Sin^3B = 3SinB.\\ Using Sin Law, \dfrac{AC}{SinB}=\dfrac{AB}{Sin(180-3B)} .\\ But\ Sin(180-3B)=Sin3B=- 4Sin^3B + 3SinB=SinB*(-Sin^2B\ +\ 3),\\ \therefore\ 25=\dfrac{11}{-4Sin^2B+3}.\\$ $Simplifying\ Sin^2B=\frac 2 {11}.\ \ \therefore\ cos^2B=\frac 9 {11}.\\ \therefore\ Sin2B=2*(\sqrt{\frac 2 {11}*\frac 9 {11}}=\dfrac{6\sqrt2}{11}.\\$ $Using\ Sin\ Law, BC=Sin2B*\dfrac{AC}{SinB}=\dfrac{6\sqrt2}{11}*\dfrac{11}{\sqrt{\dfrac 2 {11}}}\\ BC^2=6^2*11=396.$

Angle A opposite side a and Angle B opposite side b, all triangles (integer or not) with B=2A have a(a+c)=b^2.

11(11+25)= b^2.

b^2= 396

This can be solved very easily with strong basics. I am not sure about any theorems and shortcuts.

Draw the triangle with side AB as the base and drop a altitude from C to line AB to meet at Point P. Also draw a line from C to PB to meet at Q such that Angle CAQ = angle CQA. Inferences : 1) AP =PQ , AC= CQ=11

2) We know Angle ABC = x ,Angle CAB = 2x, so angle ACB =180-3x.

3) Angle ACQ = 180-4x (Angle CAQ = angle CQA).

4) angle QCB = x , [(180-3x - (180-4x))], so CQ=QB=11.

BC^{2} = CP^{2} +PB^{2}

RHS can be written as

= AC^{2} - AP^{2}+PB^{2}

= AC^{2} - [AP^{2} - PB^{2}]

= AC^{2} - [ (AP - PB)*(AP+PB) ]

= AC^{2} - [ (AP - PB)*(AB) ]

= AC^{2} + [ (PB - AP )*(AB) ]

= AC^{2} + [ (PB - PQ )*(AB) ] .......................... (AP=PQ)

= AC^{2}+ [ (QB )*(AB) ]

= 11 11+[ 11 25] ................ eventually leading to Stewart's theorem (mentioned by Aryan.C)

= 396