Expanding Triangle

First of all, we see that $\left( AB \right) BC=\frac { 1740 }{ 12 } =145$ . Using Law of Cosines in the first $\triangle ABC$ , we get that ${ AC }^{ 2 }={ AB }^{ 2 }+{ BC }^{ 2 }-2(AB\cdot BC)(cos\quad 60)={ AB }^{ 2 }+{ BC }^{ 2 }-290(0.5)={ AB }^{ 2 }+{ BC }^{ 2 }-145$ So, ${ 12 }^{ 2 }={ AB }^{ 2 }+{ BC }^{ 2 }-145$ Then, ${ 289={ AB }^{ 2 }+{ BC }^{ 2 } }$ We use that in the new $\triangle ABC$ , and we get that ${ AC }^{ 2 }=289={ AB }^{ 2 }+{ BC }^{ 2 }$ So, $\boxed{ AC=17 }$

The length we want to find can be written as $\sqrt{a^2+c^2}$ as it is a right angled triangle. Now the cosine rule states that $b^2=a^2+c^2-2ac\cos B$ therefore by rearranging we have an expression describing the first triangle $\sqrt{b^2+2ac\cos B}=\sqrt{a^2+c^2}$ So by simply plugging in the values for $b$ and B we have the unknown quantity $ac$ . The product of the side lengths of the first triangle is $1740$ therefore by dividing by $12$ we get $145=ac$ finishing the formula plugging in we get our answer to be $\boxed{17}$ .