Two ships A & B are anchored near the shore. From the two observation poles, one on each ship,an airplane is observed the distance from the observation pole in A to the airplane is 400 ft while distance from the observation pole B to the airplane,the distance is 300 m. If the angle in A is 45°, Find The distance bet. The two ships? 7. Law of Cosine (Solve the Triangle): a=25, b = 10 and c = 20
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Answer:
In the first scenario, let's solve for the distance between the two ships, given the distance from observation pole A to the airplane (400 ft), the distance from observation pole B to the airplane (300 m), and the angle at observation pole A (45°).
Using the Law of Cosines:
c^2 = a^2 + b^2 - 2ab*cos(C)
Where:
c = distance between the two ships (unknown)
a = distance from observation pole A to the airplane (400 ft)
b = distance from observation pole B to the airplane (300 m)
C = angle at observation pole A (45°)
Converting 300 m to feet (1 m ≈ 3.281 ft):
b = 300 m * 3.281 ft/m = 984.3 ft
Plugging in the values into the equation:
c^2 = (400 ft)^2 + (984.3 ft)^2 - 2 * (400 ft) * (984.3 ft) * cos(45°)
Simplifying:
c^2 = 160,000 ft^2 + 968,176.49 ft^2 - 787,424 ft^2
c^2 = 340,752.49 ft^2
Taking the square root of both sides:
c ≈ √(340,752.49 ft^2) ≈ 584.2 ft
Therefore, the distance between the two ships is approximately 584.2 feet.
Now, moving on to the second scenario, where we need to solve the triangle with side lengths a = 25, b = 10, and c = 20 using the Law of Cosines.
c^2 = a^2 + b^2 - 2ab*cos(C)
Plugging in the values:
(20)^2 = (25)^2 + (10)^2 - 2 * (25) * (10) * cos(C)
400 = 625 + 100 - 500 * cos(C)
400 = 725 - 500 * cos(C)
500 * cos(C) = 725 - 400
500 * cos(C) = 325
cos(C) = 325 / 500
cos(C) ≈ 0.65
C ≈ cos^(-1)(0.65)
C ≈ 48.19°
Therefore, the angle C is approximately 48.19°.