Answer:
The angle of elevation of the top of a building from the foot of the tower is 30°.
The angle of elevation of the top of tower from the foot of the building is 60°.
Height of the tower is 50 m.
Let the height of the building be h.
Step-by-step explanation:
[tex]\sf In \: \Delta BDC \\ [/tex]
[tex]:\implies \sf tan \: \theta = \dfrac{Perpendicular}{Base} \\ \\ [/tex]
[tex]:\implies \sf tan \: 60^{\circ} = \dfrac{CD}{BD} \\ \\ [/tex]
[tex]:\implies \sf \sqrt{3} = \dfrac{50}{BD} \\ \\ [/tex]
[tex]:\implies \sf BD = \dfrac{50}{ \sqrt{3} } \: m\\ \\ [/tex]
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[tex]\sf In \: \Delta ABD, \\ [/tex]
[tex]\dashrightarrow\:\: \sf tan \: 30 ^{\circ} = \dfrac{AB}{BD} \\ \\ [/tex]
[tex]\dashrightarrow\:\: \sf \dfrac{1}{ \sqrt{3} } = \dfrac{AB}{BD} \\ \\ [/tex]
[tex]\dashrightarrow\:\: \sf \dfrac{1}{ \sqrt{3} } = \dfrac{h}{ \dfrac{50}{ \sqrt{3} } } \\ \\ [/tex]
[tex]\dashrightarrow\:\: \sf h = \dfrac{ 50 }{ \sqrt{3} \times \sqrt{3} } \\ \\ [/tex]
[tex]\dashrightarrow\:\: \underline{ \boxed{\sf Height= \dfrac{50}{3} \: m }}\\ \\ [/tex]
[tex]\therefore\:\underline{\textsf{Height of the building is \textbf{50/3 meter}}}. \\ [/tex]
