The shadow of a vertical tower on level ground increases by 16 m when the altitude of the sun changes - Maths - Some Applications of Trigonometry

Question

The shadow of a vertical tower on level ground increases by 16 m
when the altitude of the sun changes from angle of elevation 60
degree to 45 degree Find the height of the tower,correct to one
place of decimal (Take root 3=1 73)

Chetna Khanna · Accepted Answer

Let AB be the vertical tower Let its height be h m CD = 16 m (given) Let BC = x m In Î”ABC, $\tan 60 ° = \frac{AB}{BC} ∴ \sqrt{3} = \frac{h}{x} \Rightarrow h = x \sqrt{3} (1)$ In Î”ABD, $\tan 45 ° = \frac{AB}{BD} ∴ 1 = \frac{h}{x + 16} \Rightarrow h = x + 16 \Rightarrow x \sqrt{3} = x + 16 [from (1)] \Rightarrow x (\sqrt{3} - 1) = 16$ $\Rightarrow x = \frac{16}{(\sqrt{3} - 1)} h = \frac{16}{(\sqrt{3} - 1)} \sqrt{3} = \frac{16 \sqrt{3} (\sqrt{3} + 1)}{(\sqrt{3} - 1) (\sqrt{3} + 1)} = \frac{16 \sqrt{3} (\sqrt{3} + 1)}{(3 + 1)}$ $= \frac{16 (3 + \sqrt{5})}{2} = 8 (3 + \sqrt{5}) = 8 (3 + 1 7321) = 8 \times 4 7321 = 37 86 (approx) = 37 9$ So, the height of tower is 37 9 m (approx)

The shadow of a vertical tower on level ground increases by 16 m when the altitude of the sun changes from angle of elevation 60 degree to 45 degree.Find the height of the tower,correct to one place of decimal.(Take root 3=1.73)