Answer:
10701.97 feet.
Step-by-step explanation:
Refer the attached figure
The angle of elevation of a point, A, at the top of a cliff is 21° i.e. ∠ACB = 21°
The ship has sailed 2,500 feet directly toward the foot of the cliff i.e. CD = 2500 feet.
Then the angle of elevation becomes 47° i.e.∠ADB = 47°
Let BD be x
So, BC = BD+DC=x+2500
Let the height of the cliff be h feet.
In ΔABD
We will use trigonometric ratios
[tex]tan\theta = \frac{Perpendicular}{Base}[/tex]
[tex]tan47^{\circ} = \frac{AB}{BD}[/tex]
[tex]1.072= \frac{h}{x}[/tex]
[tex]1.072x=h[/tex] ---a
In ΔABC
We will use trigonometric ratios
[tex]tan\theta = \frac{Perpendicular}{Base}[/tex]
[tex]tan21^{\circ} = \frac{AB}{BC}[/tex]
[tex]0.869 = \frac{h}{x+2500}[/tex]
[tex]0.869(x+2500) =h[/tex]
[tex]0.869x+2172.5 =h[/tex] -----b
Equate a and b
[tex]0.869x+2172.5 =1.072x[/tex]
[tex]2172.5 =1.072x-0.869x[/tex]
[tex]2172.5 =0.203x[/tex]
[tex]\frac{2172.5}{0.203}=x[/tex]
[tex]10701.97=x[/tex]
Thus the height of the cliff is 10701.97 feet.
