City planners are creating a 3-dimensional model of a city park to find the best placement for a new tennis court. The city park
1 answer:
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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
