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KatRina [158]
3 years ago
14

Jack looks at a clock tower from a distance and determines that the angle of elevation of the top of the tower is 40°. John, who

is standing 20 meters from Jack as shown in the diagram, determines that the angle of elevation to the top of the tower is 60°. If Jack’s and John’s eyes are 1.5 meters from the ground and the distance from Jack's eyes to the top of the tower is 50.64 feet, how far is John from the base of the tower?
24.5 meters
16.1 meters
22.2 meters
18.8 meters

Mathematics
2 answers:
gtnhenbr [62]3 years ago
7 0

Answer:

Option 4 - 18.8 meter.

Step-by-step explanation:

Given : Jack looks at a clock tower from a distance and determines that the angle of elevation of the top of the tower is 40°. John, who is standing 20 meters from Jack.

Determines that the angle of elevation to the top of the tower is 60°. If Jack’s and John’s eyes are 1.5 meters from the ground and the distance from Jack's eyes to the top of the tower is 50.64 feet.

To find : How far is John from the base of the tower?

Solution :

Let x be the distance between John and clock tower.

Let y be the vertical distance from the eyes of the two men  standing to the top of the clock tower.

First we take a right angle triangle ABD,

Apply trigonometric,

\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}

\tan(40)=\frac{(y + 1.5)}{(x + 20)}

\tan(40)\times (x+20)=(y + 1.5)            

y=[\tan(40)\times (x+20)]-1.5      

Now, we take right angle triangle ACD,

\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}

\tan(60)=\frac{(y + 1.5)}{(x)}

\tan(60)\times (x)=(y + 1.5)            

y=[\tan(60)\times x]-1.5      

Now, Solving for x equating both y

[\tan(60)\times x]-1.5=[\tan(40)\times (x+20)]-1.5    

\tan(60)\times x=\tan(40)\times (x+20)    

\tan(60)\times x=\tan(40)\times x+\tan(40)\times 20)    

x(\tan(60)-\tan(40))=20\tan(40)    

x=\frac{20\tan(40)}{\tan(60)-\tan(40)}    

x=\frac{20(0.8390)}{1.732050-0.8390}    

x=\frac{16.7819}{0.8929}    

x=18.79              

Therefore, Option 4 is correct.

Distance of John from the base of the tower is 18.8 meter.

Refer the attached figure.

Margaret [11]3 years ago
6 0

tan(40) = (y + 1.5) / (x + 20)

y + 1.5 = (x + 20)[tan(40)]

y = (x + 20)[tan(40)] - 1.5

tan(60) = (y + 1.5) / x

y + 1.5 = (x)[tan(60)]

y = (x)[tan(60)] - 1.5

(x)[tan(60)] - 1.5 = (x + 20)[tan(40)] - 1.5

(x)[tan(60)] = (x + 20)[tan(40)]

(x)[tan(60)] = (x)[tan(40)] + (20)[tan(40)]

(x)[tan(60)] - (x)[tan(40)] = (20)[tan(40)]

(x)[tan(60) - tan(40)] = (20)[tan(40)]

x = [(20)(tan(40))] / [tan(60) - tan(40)]

x = 18.8 meters

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