Question

Hari, standing on the top of a building, sees the top of a tower at an angle of elevation of 50c and the foot of the tower at an angle of depression of 20c. Hari is 1.6 metre tall and the height of the building on which he is standing is 9.2 mitres. [3] (a) Draw a rough sketch according to the given information. (b) How far is the tower from the building? (c) Calculate the height of the tower. [sin 20c= 0 34 . , cos 20c= 0 94 . , tan 20c= 0 36 . sin 50c= 0 77 . , cos 50 0 64 c = . , tan 50c= 1 19 . ]

Answer 1

The tower is 30 m away from the building. The height of the tower is 46.5 m.

Step-by-step explanation:

Referring to the figure attached below let’s make certain assumptions:

Height of the building, BC = 9.2 m

Hari’s height, AB = 1.6 m

The angle of elevation of the top of the tower, ∠GAF = 50°

The angle of depression of the foot of the tower, ∠FAD = ∠ADC = 20° … [since AF // CD]

Since AF // BE // CD, then let the distance between the tower and the building be CD = AF = “x” m.

Also, AB // FE, ∴ AB = FE = 1.6 m and BC // ED, ∴ BC = ED = 9.2 m

Let the height of the tower, GD = GF + FE + ED

Step 1:

Consider ∆ ACD, applying the trigonometry property of the triangle, we get

tan θ = $\frac{Perpendicular}{Base} = \frac{AC}{CD}$

⇒ tan 20° = $\frac{1.6 + 9.2}{x}$

⇒ 0.36 = $\frac{10.8}{x}$ …… [tan 20° = 0.36 given]

⇒ x = $\frac{10.8}{0.36}$

⇒ x = CD = AF = 30 m  

Thus, the tower is 30 m away from the building.

Step 2:

Consider ∆ AGF, applying the trigonometry property of the triangle, we get

tan θ = perpendicular/base = $\frac{Perpendicular}{Base} = \frac{GF}{AF}$

⇒ tan 50° = $\frac{GF}{30}$

⇒ 1.19 = $\frac{GF}{30}$ …… [tan 50° = 1.19 given]

⇒ GF = 1.19 * 30

⇒ GF = 35.7 m

Step 3:

Thus,  

The height of the tower "GD" is given as,

= 35.7 + 1.6 + 9.2

= 46.5 m

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Answer 2

Answer:

Step-by-step explanation:

The tower is 30 m away from the building. The height of the tower is 46.5 m

Height of the building, BC = 9.2 m

Hari’s height, AB = 1.6 m

The angle of elevation of the top of the tower, ∠GAF = 50°

The angle of depression of the foot of the tower, ∠FAD = ∠ADC = 20° … [since AF // CD]

mool

Since AF // BE // CD, then let the distance between the tower and the building be CD = AF = “x” m.

Also, AB // FE, ∴ AB = FE = 1.6 m and BC // ED, ∴ BC = ED = 9.2 m

Let the height of the tower, GD = GF + FE + ED

:

Consider ∆ ACD, applying the trigonometry property of the triangle, we get

tan θ = \frac{Perpendicular}{Base} = \frac{AC}{CD}

Base

Perpendicular

​

=

CD

AC

​

⇒ tan 20° = \frac{1.6 + 9.2}{x}

x

1.6+9.2

​

⇒ 0.36 = \frac{10.8}{x}

x

10.8

​

…… [tan 20° = 0.36 given]

⇒ x = \frac{10.8}{0.36}

0.36

10.8

⇒ x = CD = AF = 30 m

, the tower is 30 m away from the building.

Consider ∆ AGF, applying the trigonometry property of the triangle, we get

tan θ = perpendicular/base = \frac{Perpendicular}{Base} = \frac{GF}{AF}

Base

Perpendicular

​

=

AF

GF

​

⇒ tan 50° = \frac{GF}{30}

30

GF

​

⇒ 1.19 = \frac{GF}{30}

30

GF

​

…… [tan 50° = 1.19 given]

⇒ GF = 1.19 * 30

⇒ GF = 35.7 m

The height of the tower "GD" is given as,

= 35.7 + 1.6 + 9.2

= 46.5 m