The length of the shadow of a vertical tower on level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. The height of the tower is?
Answer & explanation
Correct answer: option 1
In ΔABC;
tan 45° = 1 : 1
(AB) (BC) ⇒ AB = BC = x
In ΔABD;
tan 30° = 1 : \(\sqrt {3}\)
(AB) (BC)
↓ ↓
x \(\sqrt {3}\)x
Now;
CD = BD - CD = \(\sqrt {3}\)x - x
⇒ (\(\sqrt {3}\) - 1)x = 10
⇒ x = \(\frac{10}{\sqrt {3} - 1}\) = \(\frac{10 (\sqrt {3} + 1)}{2}\) = 5(\(\sqrt {3}\) + 1)
AB = x = 5(\(\sqrt {3}\) + 1) m