Description for Correct answer: Let the height of the tower be h. \( \Large In \triangle BCD, \tan 45 ^{\circ} = \frac{h}{BC} \) => BC = h \( \Large In \triangle ABD, \tan 30 ^{\circ} = \frac{h}{AB} \) => \( \Large AB = \frac{h}{\tan 30 ^{\circ} } \) => \( \Large AB= \sqrt{3}h => AC+BC=\sqrt{3}h \) => \( \Large 60 + h = \sqrt{3}h \) [using (i)] => \( \Large h = \frac{60}{\sqrt{3}-1} = \frac{60 \left(\sqrt{3}+1\right) }{2} = 30 \left(\sqrt{3}+1\right) m \) Part of solved Height and Distance questions and answers : >> Elementary Mathematics >> Height and Distance Comments Similar Questions
11. An aeroplane flying horizontally at a height of 3 Km, above the ground is observed at a certain point on earth to subtend an angle of 60°. After 15 sec flight, its angle of elevation is changed to 30°. The speed of the aeroplane (taking √3 = 1.732) is
B. 230.93 m/sec AB = CD = 3000 metreA and C = Positions of aeroplane In ΔOCD, ∴ Speed of aeroplane
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