2013 NS 29
1.
cos-theorem:
\(\begin{array}{|rcll|} \hline \mathbf{ b^2 } &\mathbf{=}& \mathbf{d^2+(a-c)^2-2d(a-c)\cos(L)} \\\\ 65^2&=& 60^2+58^2-120\cdot 58\cos(L) \\ \cos(L)&=& \dfrac{60^2+58^2-65^2 }{120\cdot 58} \\ \cos(L)&=& \dfrac{2739}{6960} \\ \cos(L)&=& 0.39353448276 \\ L &=& \arccos(0.39353448276) \\ L &=& 66.8253951955^\circ \\ \mathbf{\sin( L)} &\mathbf{=}& \mathbf{0.91930985575} \\ \hline \end{array}\)
2.
cos-theorem:
\(\begin{array}{|rcll|} \hline \mathbf{ d^2 } &\mathbf{=}& \mathbf{b^2+(a-c)^2-2b(a-c)\cos(K)} \\\\ 60^2&=& 65^2+58^2-130\cdot 58\cos(K) \\ \cos(K)&=& \dfrac{65^2+58^2-60^2 }{130\cdot 58} \\ \cos(K)&=& \dfrac{3989}{7540} \\ \cos(K)&=& 0.52904509284 \\ K &=& \arccos(0.52904509284 ) \\ K &=& 58.0590417221^\circ \\ \mathbf{\sin(K)} &\mathbf{=}& \mathbf{0.84859371300} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline 1. & \sin(K) &=& \dfrac{r}{x} \qquad \text{ or } \qquad r = x\sin(K) \\\\ 2. & \sin(L) &=& \dfrac{r}{80-x} \quad | \quad r = x\sin(K) \\ & \sin(L) &=& \dfrac{x\sin(K)}{80-x} \\ & (80-x)\sin(L) &=& x\sin(K) \\ & 80\sin(L)-x\sin(L) &=& x\sin(K) \\ & x\sin(L)+x\sin(K) &=& 80\sin(L) \\ & x\Big(\sin(L)+\sin(K)\Big) &=& 80\sin(L) \\ & \mathbf{x} &\mathbf{=}& \mathbf{80\cdot \dfrac{ \sin(L)} {\sin(L)+\sin(K)} }\\\\ & x & = & 80\cdot \dfrac{0.91930985575} {0.91930985575+0.84859371300} \\ & x & = & 80 \cdot 0.52 \\ & x & = & 41.6 \\ & \mathbf{x} &\mathbf{=}& \mathbf{41 \ \dfrac{3}{5}} \\ \hline \end{array}\)
The length of segment KA is \(\mathbf{41 \ \dfrac{3}{5}}\)
