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From a point on the ground, the angles of elevation of the bottom and top of a tower fixed at the top of a \( 30 m \) high building are \( 45^{\circ} \) and \( 60^{\circ} \) respectively. Find the height of the tower. \( (\sqrt{3}=1.732) \)

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In right ΔBAD,

tan 45° = \(\frac{AB}{AD}\)

⇒ \(\frac{AB}{AD}\) = 1 (\(\because\) tan 45° = 1)

⇒ AD = AB = 30 m (\(\because\) AB = Height of house = 30 m)

In right Δ CAD,

tan 60° = \(\frac{AC}{AD}\) 

⇒ AC = AD\(\sqrt3\) (tan 60° = \(\sqrt3\))

 = 30\(\sqrt3\).

⇒  AB + BC = 30\(\sqrt3\)

⇒ BC = 30\(\sqrt3\) - AB = (30\(\sqrt3\) - 30)m

⇒ BC = 30(\(\sqrt3\) - 1)m

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