From the top of a cliff 25m high the angle of elevation of a tower is found to be equal to the angle

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This is the diagram of the given question. CD is the cliff of height 25 m and EF is the tower with base E and top F. Let $\theta $ be the angle of elevation and depression. From the given figure we can see that-CD = EG = 25 m ….(1)Applying trigonometric ratios in triangles DGE and DGF, $\begin{gathered} tan\theta = \dfrac{{EG}}{{DG}} \\ tan\theta = \dfrac{{FG}}{{DG}} \\ \end{gathered} $By comparing the two values we can see that,$\dfrac{{EG}}{{DG}} = \dfrac{{FG}}{{DG}}$EG = FGUsing equation (1), we can see thatEG = FG = 25 mThe height of the tower is EF = EG + GFEF = 25 + 25 EF = 50 mThis is the required answer. Note: In such types of questions, it is important to read the language of the question carefully and draw the diagram step by step correctly. When the diagram is drawn, we just have to apply basic trigonometry to find the required 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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