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
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,
[tex]\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}[/tex]
[tex]\tan(40)=\frac{(y + 1.5)}{(x + 20)}[/tex]
[tex]\tan(40)\times (x+20)=(y + 1.5)[/tex]
[tex]y=[\tan(40)\times (x+20)]-1.5[/tex]
Now, we take right angle triangle ACD,
[tex]\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}[/tex]
[tex]\tan(60)=\frac{(y + 1.5)}{(x)}[/tex]
[tex]\tan(60)\times (x)=(y + 1.5)[/tex]
[tex]y=[\tan(60)\times x]-1.5[/tex]
Now, Solving for x equating both y
[tex][\tan(60)\times x]-1.5=[\tan(40)\times (x+20)]-1.5[/tex]
[tex]\tan(60)\times x=\tan(40)\times (x+20)[/tex]
[tex]\tan(60)\times x=\tan(40)\times x+\tan(40)\times 20)[/tex]
[tex]x(\tan(60)-\tan(40))=20\tan(40)[/tex]
[tex]x=\frac{20\tan(40)}{\tan(60)-\tan(40)}[/tex]
[tex]x=\frac{20(0.8390)}{1.732050-0.8390}[/tex]
[tex]x=\frac{16.7819}{0.8929}[/tex]
[tex]x=18.79[/tex]
Therefore, Option 4 is correct.
Distance of John from the base of the tower is 18.8 meter.
Refer the attached figure.
