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The angles of depression of two ships from the top of a light house are 60° and 45° towards east. If the ships are 300 metre apart, the height of the light house is
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- 200 (3 + √3) metre
- 250 (3 + √3) metre
- 150 (3 + √3) metre
- 160 (3 + √3) metre
Correct Option: C

AB = Lamp post = h metre C and D = Positions of ships CD = 300 metre; BC = x metre ∆ACB = 60 metre; ∠ADB = 45°
In ∆ABC,
| tan 60° = | |
| BC |
| ⇒ √3 = | |
| x |
⇒ h = √3x ..... (i)
In ∆ABD,
| tan 45° = | |
| BD |
| ⇒ 1 = | |
| x + 300 |
⇒ h = x + 300
| ⇒ h = | + 300 | |
| √3 |
| ⇒ h - | = 300 | |
| √3 |
| ⇒ h - | = 300 | |
| √3 |
⇒ h (√3 - 1) = 300√3
| ⇒ h = | |
| √3 - 1 |
| = | |
| (√3 - 1)(√3 + 1) |
| = | |
| 2 |
= 150(3 + √3) metre
= 45 kmph.