From the top of a cliff,60 metres high, the angles of depression of the top and bottom of a tower are observed to be 30 ° and 60 °.Find the height of the tower.

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From the top of a cliff,60 metres high, the angles of depression of the top and bottom of a tower are observed to be 30 ° and 60 °.Find the height of the tower.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I can't understand..
\[\tan\theta = \frac{\text{opp}}{\text{adj}}\]
Let AB be the building h meters high, and PQ be the tower 60 m high situated at distance BQ = x meters away. Then AC = BQ = x, CQ = AB = h, PC = 60 -h From BPQ, cot 60° = BQ/PQ = x/60 => x = 60 cot 60° From APC, cot 30° = AC/PC = x/(60 -h) => 60 -h = x/ cot 30° = 60 cot 60° tan 30° = 60 (1/3)(1/3) = 20 => h = 60 -20 = 40 Thus the building is 40 meters high and is situated 34·64 meters away from the tower.

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i've given you the figure pls refer to it
^^^^ i think the use of inverse trigonometric functions is unnecessary.
no by cot theta it is helpful and cot theta is just the opposite of tan theta
Yes, but I don't think this person has actually learnt inverse trigonometry, and for the sake of simplicity I don't think he/she needs to know it for now.

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