Question

A man, m tall, originally stands at a distance of m from a building whose height above the level ground is Hm. At this position, he sights the top of the building at an elevation angle of 45. He then walks 1 m closer to the building, and observed that the elevation angle became 60. Prove that: 1) H=x+y
2) H=x+tan60(y-1)

Answer 1

Given:

Height of the man = x m

Height of the building = H m

The man is y m away from the building

The angle of elevations from the initial position is 45°

The angle of elevation after walking 1 m closer to the building is 60°

To find:

(1). H = x + y

(2). H = x + tan 60(y - 1)

Solution:

The following trigonometric function of a triangle will be used to solve the given problem:  

 $\boxed{\bold{ tan \:\theta = \frac{Opposite\:Side}{Adjacent\:Side} }}$

Referring to the figure attached below, we have,

AB = H m = Height of the building

EF = DG = x m = Height of the man

FG = ED = 1 m = distance walked by the man

FB = EC = y m = distance between the initial position of the man and the building

Case (1): Proving H = x + y:

Consider ΔAEC, we have

θ = 45°

Opposite side = AC = (H - x) m

Adjacent side = FB = y m

Now, substituting the values in the trigonometric function mentioned above,

$tan \:45 = \frac{AC}{EC}$

⇒ $tan \:45 = \frac{H \:-\: x}{y}$

⇒ $1 = \frac{H \:-\: x}{y}$

⇒ $y = H\:-\:x$

⇒ $\boxed{\bold{H = x\:+\:y}}$

Hence proved

Case (2): Proving H = x + tan60(y-1):

Consider ΔADC, we have

θ = 60°

Opposite side = AC = (H - x) m

Adjacent side = DC = BG = (y - 1) m

Now, substituting the values in the trigonometric function mentioned above,

$tan \:60 = \frac{AC}{DC}$

⇒ $tan \:60 = \frac{H \:-\: x}{y\:-\:1}$

⇒ $(y - 1)tan\:60 = H\:-\:x$

⇒ $\boxed{\bold{H = x \:+\:tan\:60\:(y - 1)}}$

Hence proved

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