Answer to Question #99033 in Trigonometry for Lakpayanji

Question

A rope fancer was walking on a loose rope tied to the tops of two posts each 9m high.when the dancer was 2.7manive the ground it was found that the stretched pieces ig rope made angles of 30 and 60 with horizontal line parallel to ground find the length of rope in stretched condition

Answer 1

$a_{1}=9-2.7=6.3m; \\ b_{1}=\frac{l_{1}}{2}; \\ l^2_{1}=a^2_{1}+b^2_{1}; \\ l^2_{1}=a^2_{1}+(\frac{l_{1}}{2})^2=a^2_{1}+\frac{l^2_{1}}{4}; \\ 3\cdot l^2_{1}=4\cdot a^2_{1}; \\ l_{1}=\sqrt{\frac{4\cdot a^2_{1}}{3}}=\frac{2\cdot a_{1}}{\sqrt{3}}=\frac{2\cdot 6.3}{\sqrt{3}}; \\ l_{1}=7.3m; \\ a_{2}=\frac{l_{2}}{2}; \\ l_{2}=2\cdot a_{2}; \\ a_{2}=9-2.7=6.3 m;\\ l_{2}=2\cdot 6.3=12.6 m; \\ l = l_{1}+l_{2}=7.3+12.6; \\ \textbf{l=19.9m}$