A plane is located at C on the diagram. There are two towers located at A and B. The distance between the towers is 7,600 feet, and the angles of elevation are given.
a. Find BC, the distance from Tower 2 to the plane, to the nearest foot.
b. Find CD, the height of the plane from the ground, to the nearest foot.
Tower 1:
Base angle = 16°
Tower 2:
Base angle = 24°
Base angle (for both towers) = 7600 ft. distance
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To find the distance from Tower 2 to the plane (BC), we can use trigonometry, specifically the tangent function.
In this case, we have the base angle of Tower 2 (24°) and the distance between the towers (7,600 feet).
Using the tangent function, we can set up the following equation:
tan(24°) = BC / 7600 feet
To solve for BC, we can rearrange the equation as follows:
BC = tan(24°) * 7600 feet
Now we can calculate BC using a calculator or a computer program:
BC ≈ tan(24°) * 7600 ≈ 3395.85 feet
Therefore, the distance from Tower 2 to the plane (BC) is approximately 3,396 feet.
To find the height of the plane from the ground (CD), we first need to find the distance from Tower 1 to the plane (AC). Using the same approach as above, we can use the base angle of Tower 1 (16°):
tan(16°) = AC / 7600 feet
Rearranging the equation:
AC = tan(16°) * 7600 feet
Calculating AC:
AC ≈ tan(16°) * 7600 ≈ 2151.12 feet
Now, to find the height CD, we can use the given diagram. CD is the vertical distance from the plane to the ground, and it is a right triangle (using the right angle at C).
The length of the hypotenuse AD (which is also the distance from Tower 1 to the plane) is AC + BC:
AD = AC + BC ≈ 2151.12 feet + 3395.85 feet
AD ≈ 5546.97 feet
With AD and BC known, we can use the Pythagorean theorem to find CD:
CD^2 = AD^2 - BC^2
CD^2 ≈ 5546.97^2 - 3395.85^2
CD^2 ≈ 30796198.35 - 11528616.35
CD^2 ≈ 19267582
Taking the square root of both sides:
CD ≈ √19267582 ≈ 4390.26 feet
Therefore, the height of the plane from the ground (CD) is approximately 4,390 feet.
To find BC, the distance from Tower 2 to the plane, we can use trigonometry.
a. Find BC:
Using trigonometry, we can use the tangent function to find BC. The tangent of an angle is equal to the opposite side divided by the adjacent side.
Let's assume BC = x.
In triangle BCD, we have:
tan(24°) = x / 7600 ft
Rearranging the equation:
x = tan(24°) * 7600 ft
Using a calculator, tan(24°) ≈ 0.44504187.
Substituting the value into the equation:
x ≈ 0.44504187 * 7600 ft
Calculating the result:
x ≈ 3386.320048 ft
Therefore, BC, the distance from Tower 2 to the plane, is approximately 3386 ft (to the nearest foot).
b. Find CD:
To find the height of the plane from the ground, CD, we can use trigonometry again.
In triangle ACD, we have:
tan(16°) = CD / 7600 ft
Rearranging the equation:
CD = tan(16°) * 7600 ft
Using a calculator, tan(16°) ≈ 0.28674539.
Substituting the value into the equation:
CD ≈ 0.28674539 * 7600 ft
Calculating the result:
CD ≈ 2176.789744 ft
Therefore, CD, the height of the plane from the ground, is approximately 2177 ft (to the nearest foot).