Answer:
0.67 mi
Step-by-step explanation:
The diagram illustrating the question is shown in the attach photo.
In triangle DCA,
Opposite = H
Adjacent = b
Angel θ = 27°
Tan θ = Opp /Adj
Tan 27° = H/b
Cross multiply
H = b x Tan 27°... (1)
From triangle DSA,
The diagram illustrating the question is shown in attach photo.
In triangle DCA,
Opposite = H
Adjacent = 2.3 – b
Angel θ = 34°
Tan θ = Opp /Adj
Tan 34° = H/ 2.3 – b
Cross multiply
H = Tan 34° (2.3 – b) .. (2)
Equating equation (1) and (2)
b x Tan 27° = Tan 34° (2.3 – b)
0.5095b = 0.6745(2.3 – b)
0.5095b = 1.55135 – 0.6745b
Collect like terms
0.5095b + 0.6745b = 1.55135
1.184b = 1.55135
Divide both side by 1.184
b = 1.55135/1.184
b = 1.31 mi
Substitute the value of b into any of the equation to obtain the height (H). In this case we shall use equation 1.
H = b x Tan 27°
H = 1.31 x Tan 27°
H = 0.67 mi
Therefore, the height of the drone is 0.67 mi
The height of the drone is 0.67 mi
In triangle DCA,
Opposite = H
Adjacent = b
Angel θ = 27°
Tan θ = [tex]Opp \div Adj[/tex]
Tan 27° =[tex]H\div b[/tex]
Now we have to do the cross multiply
[tex]H = b \times Tan 27^{\circ}... (1)[/tex]
Now
From triangle DSA,
In triangle DCA,
Opposite = H
Adjacent = 2.3 – b
Angel θ = 34°
Tan θ = [tex]Opp \div Adj[/tex]
Tan 34° = [tex]H\div 2.3 - b[/tex]
Now we have to do the cross multiply
H = Tan 34° (2.3 – b) .. (2)
Now
Equating equation (1) and (2)
[tex]b \times Tan 27^{\circ} = Tan 34^{\circ} (2.3 - b)[/tex]
0.5095b = 0.6745(2.3 – b)
0.5095b = 1.55135 – 0.6745b
Now
0.5095b + 0.6745b = 1.55135
1.184b = 1.55135
b = 1.31 mi
Now
[tex]H = b \times Tan 27^{\circ}\\\\H = 1.31 \times Tan 27^{\circ}[/tex]
H = 0.67 mi
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