Answer:
The length of the antenna is 174 feet.
Step-by-step explanation:
Given that the horizontal distance from the bottom of the radio aAntenna is 600 feet.
The angle of elevation to the bottom of the antenna is 10∘, and the angle of elevation to the top of the antenna is 25∘.
Let AB is the hill, BC is the antenna on the top of the hill, and point O is the viewer's eye as shown in the figure.
[tex]\angle AOB = 10^{\circ}[/tex]
[tex]\angle AOC = 25^{\circ}[/tex]
OA=600 feet
The length of the antenna = BC
From the figure, BC=AC-AB...(i)
Now in triangle AOB, we have
[tex]\tan 10^{\circ} = \frac {AB}{OA} \\\\\Rightarrow AB=OA\times \tan 10^{\circ} \\\\\Rightarrow AB=600\times \tan 10^{\circ} \\\\[/tex]
[tex]\Rightarrow AB = 105.8[/tex] feet ...(ii)
Similarly, in triangle AOC, we have
[tex]\tan 25^{\circ} = \frac {AC}{OA} \\\\\Rightarrow AC=OA\times \tan 25^{\circ} \\\\\Rightarrow AC=600\times \tan 25^{\circ} \\\\[/tex]
[tex]\Rightarrow AC = 279.8[/tex] feet ...(iii)
By applying the values from equation (ii) and (iii) in the equation (i), we have,
BC=279.8-105.8=174 feet
Hence, the length of the antenna is 174 feet.
The question is an illustration of angles of elevation and height
The height of the antenna is 174 feet.
Considering triangle ABC, we have the following tangent ratio
[tex]\tan(25) =\frac{AB}{BC}[/tex]
So, we have:
[tex]\tan(25) =\frac{x + h}{600}[/tex]
Considering triangle DBC, we have:
[tex]\tan(10) =\frac{DB}{BC}[/tex]
So, we have:
[tex]\tan(10) =\frac{x}{600}[/tex]
Make x the subject
[tex]x = 600\tan(10)[/tex]
Substitute [tex]x = 600\tan(10)[/tex] in [tex]\tan(25) =\frac{x + h}{600}[/tex]
[tex]\tan(25) =\frac{ 600\tan(10) + h}{600}[/tex]
Multiply both sides by 600
[tex]600\tan(25) =600\tan(10) + h[/tex]
Make h the subject
[tex]h = 600\tan(25) -600\tan(10)[/tex]
Evaluate tan(25) and tan(10)
[tex]h = 600\times 0.4663 -600\times 0.1763[/tex]
[tex]h = 174[/tex]
Hence, the height of the antenna is 174 feet.
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