geometry

This problem is neatly solved by Presh Talwalker on YouTube. Search for "Mind Your Decisions. The chord progression problem."

Solve for x.

\(\begin{array}{|rcll|} \hline \mathbf{\text{cos-rule:}} \\ \hline \mathbf{z^2} &=& \mathbf{10^2+12^2-2*10*12*\cos(60^\circ)} \quad | \quad \cos(60^\circ) = \dfrac{1}{2} \\ z^2 &=& 100+144-240*\dfrac{1}{2} \\\\ z^2 &=& 124 \\\\ z^2 &=& 4*31 \\\\ \mathbf{z} &=& \mathbf{2\sqrt{31}} \qquad (1) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{\text{sin-rule:}} \\ \hline \mathbf{\dfrac{\sin(\alpha)}{10}} &=& \mathbf{\dfrac{\sin(60^\circ)}{z}} \\\\ \sin(\alpha) &=& \dfrac{10\sin(60^\circ)}{z} \quad | \quad \sin(60^\circ) = \dfrac{\sqrt{3}}{2} \\\\ \sin(\alpha) &=& \dfrac{10\sqrt{3}}{2z} \\\\ \sin(\alpha) &=& \dfrac{5\sqrt{3}}{z} \quad | \quad z=2\sqrt{31} \\\\ \sin(\alpha) &=& \dfrac{5\sqrt{3}}{2\sqrt{31}} \\\\ \sin(\alpha) &=& \dfrac{5}{62}\sqrt{93} \\\\ \sin(\alpha) &=& 0.77771377105 \\\\ \alpha &=& \arcsin\left(0.77771377105\right) \\\\ \mathbf{\alpha} &=& \mathbf{51.0517244354^\circ} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{\text{sin-rule:}} \\ \hline \mathbf{\dfrac{\sin(\alpha)}{10}} &=& \mathbf{ \dfrac{\sin(\beta)}{12}} \\\\ \sin(\beta) &=& \dfrac{12}{10}\sin(\alpha) \\\\ \sin(\beta) &=& \dfrac{6}{5}\sin(\alpha) \\\\ \sin(\beta) &=& \dfrac{6}{5}*0.77771377105 \\\\ \sin(\beta) &=& 0.93325652526 \\\\ \beta &=& \arcsin\left(0.93325652526\right) \\\\ \mathbf{\beta} &=& \mathbf{68.9482755646^\circ} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{\text{sin-rule:}} \\ \hline \mathbf{\dfrac{\sin(30^\circ+\alpha)}{x}} &=& \mathbf{ \dfrac{\sin(\beta)}{12}} \\\\ x &=& \dfrac{12\sin(30^\circ+\alpha)^\circ)}{\sin(\beta)} \\\\ x &=& \dfrac{12\sin(30^\circ+51.0517244354^\circ)}{0.93325652526} \\\\ x &=& \dfrac{12\sin(81.0517244354)}{0.93325652526} \\\\ x &=& \dfrac{12*0.98782916115}{0.93325652526} \\\\ x &=& \dfrac{11.8539499338}{0.93325652526} \\\\ \mathbf{x} &=& \mathbf{12.7017059222} \\ \hline \end{array} \)

Let  the  intersection point of the three chords  = A

Let  the other end of the chord of length 10 =  B

Let the other end  of the chord of length 12 = C

Connect BC

Let the intersection of BC and chord "x"  = D

Let the other end of chord  "x"  = E

Using the Law of Cosines

BC = sqrt  [10^2 + 12^2  - 2 (10*12)cos (60°)]  =  sqrt ( 244 - 120)   = sqrt (124) = 2sqrt (31)

And because angle BAC is bisected  then  in triangle ABC

BA / CA  =  BD / CD

10/12  = BD /CD

5/6 = BD / CD

So   BD =  (5/11)* BC =  (5/11)*2sqrt (31)  = (10/11) sqrt (31)

And CD = (6/11) * BC = (6/11) * 2sqrt (31)  = (12/11)sqrt (31)

And  using the Law of Sines

sin ABC / 12  =  sin 60 / [ 2sqrt (31)]

sin ABC  =  12 sqrt (3) / [ 4sqrt (31) ]

sin ABC = 3sqrt (3) / sqrt (31)

And   using the Law of Sines once more

AD / sin ABC = BD / sin BAD

AD / [ 3sqrt (3) /sqrt (31) ]  =  (10/11)sqrt (31) / sin 30

AD sqrt (31) / [ 3sqrt (3) [ =  2 *(10/11) sqrt (31)

AD  =  60 sqrt (3)  / 11  =  (60/11)sqrt (3)

And  using the intersecting chord theorem

AD * DE =  BC * CD

(60/11)sqrt (3)  * DE  =  (10/11)sqrt (31) * (12/11)sqrt (31)

(60/11) sqrt (3) * DE  =  3720 / 121

(660/121)sqrt (3) * DE =  3720/121

DE  =  (3720/121) (121/660) / sqrt (3)

DE =  (62/11) / sqrt (3)

DE =  (62/33)sqrt (3)

So   "x"  =   AD  + DE =    sqrt(3)  ( 60/11 + 62/33)   = sqrt (3) ( 180 + 62) / 33  = 

sqrt ( 3)  (242 / 33 )  =   sqrt (3) (22/3)   =  

22 / sqrt (3)