Equal Lengths

Relevant wiki: Solving Triangles - Problem Solving - Hard

Refer to the figure above. Extend $\overline { CA }$ to $A'$ so that $A$ is the midpoint of $\overline {CA'}$ . Also, $B$ is the midpoint of $\overline {DC}$ so we can use midpoint theorem and thus, $A'D=2\cdot 10=20$ . From midpoint theorem, $\overline { AB }$ // $\overline {A'D}$ . We can do some angle chasing and label the angles in $\Delta DA'A$

In $\Delta DA'A$ , we can use sine rule and $\frac { 20 }{ \sin { 120^\circ } } =\frac { A'A }{ \sin { 15^\circ } } =\frac { DA }{ \sin { 45^\circ } }$ .

$A'A=\frac{ 20 \sin{15^\circ}} {\sin{120^\circ}} = \frac { 10(\sqrt { 6 } -\sqrt { 2 } ) }{ \sqrt { 3 } }$

As $A$ is the midpoint of $\overline { A'C}$ , $AC=A'A=\frac { 10(\sqrt { 6 } -\sqrt { 2 } ) }{ \sqrt { 3 } }$ .

Also from sine rule in $\Delta DA'A$ , $DA=\frac { 20 \sin{45^\circ}}{\sin{120^\circ}}=\frac{20 \sqrt {6}}{3}$

Now apply cosine rule on $\Delta DAC$ , ${ DC }^{ 2 }={ (\frac { 20\sqrt { 6 } }{ 3 } ) }^{ 2 }+{ (\frac { 10(\sqrt { 6 } -\sqrt { 2 } ) }{ \sqrt { 3 } } ) }^{ 2 }-2(\frac { 20\sqrt { 6 } }{ 3 } )(\frac { 10(\sqrt { 6 } -\sqrt { 2 } }{ \sqrt { 3 } } )\cos { 60^\circ }$

${DC}^{2}=\frac{6000-2400\sqrt{3}}{9}$

$DC=\pm \frac { \sqrt { 6000-2400\sqrt { 3 } } }{ 3 }$ (reject $DC=-\frac { \sqrt { 6000-2400\sqrt { 3 } } }{ 3 }$ )

$DB= \frac { \sqrt { 6000-2400\sqrt { 3 } } }{ 6 }$

$100DB=\frac { 50\sqrt { 6000-2400\sqrt { 3 } } }{ 3 } \approx 715.5$

$\Longrightarrow \lfloor 100s\rfloor=715$

Simply We Use sine rule in triangle ADB And ABC.

Call Angle ADC = x and Angle ACD = 120-x (From angle sum property)

Also Let BD= BC = y

y/sin(15) = AB/sinx

y/sin45 = AB(sin(120-x)

Divide them and solving we get value of tanx . then find sinx

and substitute in 1st equation to get y

$\frac{s}{\sin{45^\circ}}=\frac{10}{\sin{\angle BCA}}\\\angle BCA=\sin^{-1}\left(\frac{10\sin45^\circ}{s}\right)\\\frac{2}{\sin60^\circ}=\frac{AD\cdot s}{10\sin45^\circ} \\AD=\frac{20\sqrt{3}}{\sqrt{2}}\\s=\sqrt{AD^2+100-2\times AD\times10\cos15^\circ}\\s\approx7.155\\\text{Hence, }\lfloor100s\rfloor=715$

Apply sine law on $\triangle ACB$ .

$\dfrac{10}{\sin C}=\dfrac{s}{\sin 45}$ $\implies$ $\sin C = \dfrac{10 \sin 45}{s}$

Apply sine law on $\triangle ACD$ .

$\dfrac{AD}{\sin C}=\dfrac{2s}{\sin 60}$ $\implies$ $\sin C=\dfrac{AD \sin 60}{2s}$

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$\sin C = \sin C$

$\dfrac{10 \sin 45}{s}=\dfrac{AD \sin 60}{2s}$ $\implies$ $AD=\dfrac{20 \sin 45}{\sin 60}$

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Apply cosine law on $\triangle ABD$ .

$s^2=AD^2+10^2-2(AD)(10)(\cos 15)=\left(\dfrac{20 \sin 45}{\sin 60}\right)^2+100-2\left(\dfrac{20 \sin 45}{\sin 60}\right)(10)(\cos 15)\approx 51.19661283$

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$s=\sqrt{51.19661283}\approx 7.155$

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$\lfloor 100s \rfloor = 100(7.155)=\boxed{715}$

Used law of sines for traingles ADB and ABC first. And, then substituted one equation into another to get the value of 's'.