Answer
From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower
Let AB be the cliff and CD be the tower.
Then, AB = 25 m. From B draw BE ⊥ CD.
Let ∠EBD = ∠ACB = α.
`\text{Now, }\frac{\text{DE}}{\text{BE}}=\text{tan }\alpha \text{ and }\frac{\text{AB}}{\text{AC}}=\text{tan }\alpha \text{ } `
`\therefore \frac{DE}{BE}=\frac{AB}{AC}\text{ }So,\text{ }DE=AB [ ∵ BE = AC]`
∴ CD = CE + DE = AB + AB = 2AB = 50m
Concept: Heights and Distances
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