Practicing Success
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? |
5(√3 + 1) m 10 m 5(√3 - 1) m 10(√3 - 1) m |
5(√3 + 1) m |
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 |