Question
From the top of a cliff $$50\,m$$ high, the angles of depression of the top and bottom of a tower are observed to be $${30^ \circ }$$ and $${45^ \circ }.$$ The height of tower is
A.
$$50\,m$$
B.
$$50\sqrt 3 \,m$$
C.
$$50\left( {\sqrt 3 - 1} \right)m$$
D.
$$50\left( {1 - \frac{{\sqrt 3 }}{3}} \right)m$$
Answer :
$$50\left( {1 - \frac{{\sqrt 3 }}{3}} \right)m$$
Solution :
Let height of the tower be $$h\,m$$ and distance between tower and cliffbe $$x\,m.$$
$$\eqalign{ & \therefore CD = h,BD = x \cr & {\text{In }}\Delta \,ABD,\tan {45^ \circ } = \frac{{AB}}{{BD}} \cr & {\text{or }}1 = \frac{{50}}{x} \cr & x = 50\,\,\,\,.....\left( {\text{i}} \right) \cr} $$
$$\eqalign{ & {\text{In }}\Delta \,AEC \cr & \tan {30^ \circ } = \frac{{AE}}{{EC}} = \frac{{AB - EB}}{{EC}} = \frac{{AB - DC}}{{BD}}\left( {\because EB = DC,EC = BD} \right) \cr & \frac{1}{{\sqrt 3 }} = \frac{{50 - h}}{x}{\text{ or }}x = 50\sqrt 3 - h\sqrt 3 \cr & {\text{or }}50 = 50\sqrt 3 - h\sqrt 3 {\text{ or }}h\sqrt 3 = 50\sqrt 3 - 50 \cr & {\text{or }}h = \frac{{50\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} = 50\left( {1 - \frac{1}{{\sqrt 3 }}} \right) \cr & \therefore h = 50\left( {1 - \frac{{\sqrt 3 }}{3}} \right) \cr} $$
Let height of the tower be $$h\,m$$ and distance between tower and cliffbe $$x\,m.$$
$$\eqalign{ & \therefore CD = h,BD = x \cr & {\text{In }}\Delta \,ABD,\tan {45^ \circ } = \frac{{AB}}{{BD}} \cr & {\text{or }}1 = \frac{{50}}{x} \cr & x = 50\,\,\,\,.....\left( {\text{i}} \right) \cr} $$
$$\eqalign{ & {\text{In }}\Delta \,AEC \cr & \tan {30^ \circ } = \frac{{AE}}{{EC}} = \frac{{AB - EB}}{{EC}} = \frac{{AB - DC}}{{BD}}\left( {\because EB = DC,EC = BD} \right) \cr & \frac{1}{{\sqrt 3 }} = \frac{{50 - h}}{x}{\text{ or }}x = 50\sqrt 3 - h\sqrt 3 \cr & {\text{or }}50 = 50\sqrt 3 - h\sqrt 3 {\text{ or }}h\sqrt 3 = 50\sqrt 3 - 50 \cr & {\text{or }}h = \frac{{50\left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 }} = 50\left( {1 - \frac{1}{{\sqrt 3 }}} \right) \cr & \therefore h = 50\left( {1 - \frac{{\sqrt 3 }}{3}} \right) \cr} $$