# Puneet and Karim, both are standing in front of their school building. The angle of elevation of the top of the building for Puneet is 45° and that for Karim is 60°. Find the distance between Puneet and Karim, if the height of the school building is 30 m. (Take value of √3 = 1.732 and assume that the heights of Puneet and Karim are negligible.) pls give the complete answer of this very question....dont give links

$\mathrm{In}\u25b3\mathrm{ABC}\phantom{\rule{0ex}{0ex}}\mathrm{tan}45\xb0=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{30}{\mathrm{y}}=1\phantom{\rule{0ex}{0ex}}\mathrm{y}=30\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{Now}\mathrm{in}\u25b3\mathrm{ADB}\phantom{\rule{0ex}{0ex}}\mathrm{tan}60\xb0=\frac{\mathrm{AB}}{\mathrm{DB}}=\frac{30}{\mathrm{x}}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{x}=\frac{30}{\sqrt{3}}=\frac{30}{1.732}=17.32\phantom{\rule{0ex}{0ex}}\mathrm{So}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{puneet}\mathrm{and}\mathrm{karim}=\mathrm{y}-\mathrm{x}=30-17.32=12.68\mathrm{m}$

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