Answer:
Let AB is a tower. Cars are at point C and D respectively In \[\Delta \,ABC\], \[\frac{AB}{BC}=\tan \,\,30{}^\circ \] \[\frac{100}{x}=\frac{1}{\sqrt{3}}\] \[x=100\sqrt{3}\] \[=100\times 1.732\] \[=173.2\,\,m\] In \[\Delta \,ABD\], \[\frac{AB}{BD}=\tan \,\,45{}^\circ \] \[\frac{100}{y}=1\] \[y=100\,m\] Distance between two cars \[=x+y\] \[=173.2+100\] \[=273.2\,m\]
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