Question

From the top of a building 60m high, the angle of elevation and depression of the top and the foot of another building are α and β respectively. Find the height of the second building.​

Answer 1

Step-by-step explanation:

Let the line segment AB represents the height of the lamp ( where A is the top of the lamp ), ED represents the height of the building,

While C is any point on ED such that CA ║ DB and CA = DB

Thus, by the below diagram,

In triangle ECA,

Now, in triangle EDB,

Thus,

√3 EC = 60 / √3

⇒ 3 EC = 60 ⇒ EC = 20

Since, AB = CD = ED - EC = 60 - 20 = 40 meters,

Hence, the height of the lamp post is 40 meters.

Answer 2

Answer:

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Let the line segment AB represents the height of the lamp ( where A is the top of the lamp ), ED represents the height of the building,

While C is any point on ED such that CA ║ DB and CA = DB

Thus, by the below diagram,

In triangle ECA,

tan 30^{\circ}=\frac{EC}{CA}

$\frac{1}{\sqrt{3}}=\frac{EC}{CA}\implies CA =\sqrt{3} EC$

Now, in triangle EDB,

$tan 60^{\circ}=\frac{ED}{DB}=\frac{60}{CA}$

$\implies \sqrt{3}CA = 60\implies CA = \frac{60}{\sqrt{3}}$

Thus,

√3 EC = 60 / √3

⇒ 3 EC = 60 ⇒ EC = 20

Since, AB = CD = ED - EC = 60 - 20 = 40 meters,

Hence, the height of the lamp post is 40 meters.

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