Can someone help me with this math problem ?? Pls it’s for my final !
1 answer:
Answer:
221.87 feet
Step-by-step explanation:
Given that,
A 525 ft cable runs from the top of an antenna to the ground.
The angle of elevation made by the ground to the top of an antena 25°
We need to find the height of the antenna.
Using trigonometry,
Hypotenuse, H = 525 ft
θ = 25°
So,
So, the height of the antenna is equal to 221.87 feet.
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To find the distance <em>per year</em>, we'll need to find the <em>unit rate</em> of this ratio in terms of years. The word <em>unit</em> refers to the number 1 (coming from the Latin root <em>uni-</em> ); a <em>unit rate</em> involves bringing the number we're interested in down to 1 while preserving the ratio. Since we're looking for the distance the fault line moves every one year, we'll have to bring that 175 down to one, which we can do by dividing it by 175. To preserve our ratio, we also have to divide the top by 175:
We have our answer: approximately 0.14 cm or 1.4 mm per year
To find the distance <em>per year</em>, we'll need to find the <em>unit rate</em> of this ratio in terms of years. The word <em>unit</em> refers to the number 1 (coming from the Latin root <em>uni-</em> ); a <em>unit rate</em> involves bringing the number we're interested in down to 1 while preserving the ratio. Since we're looking for the distance the fault line moves every one year, we'll have to bring that 175 down to one, which we can do by dividing it by 175. To preserve our ratio, we also have to divide the top by 175:
We have our answer: approximately 0.14 cm or 1.4 mm per year