Solution
Let height of the tower is CD and A and B are the positions of the observer such that AB = 30 m.
Given that ∠DAC = 60°, ∠DBC = 75°
| From the ΔACD, | ||
| tan 60° | = |  |
 |
= | |
| DC | = | AC ----- (i) |
| From the ΔBCD, | ||
| tan 75° | = |  |
| 2 + |
= | |
| DC | = | BC(2 + ) ------ (ii) |
| From eq (i) & (ii), we get | ||
| AC |
= | BC(2 + ) |
| (AB + BC) |
= | BC(2 + ) |
| (30 + BC) |
= | BC(2 + ) (∵ AB = 30) |
| 30 |
= | BC(2 + – ) |
| BC | = | 15 |
| From eq(ii), DC | = | 15(2 + ) |
∴ Height of the tower is 15(3 + 2) m
) m