A navy cruiser detects a nuclear submarine that is 2500 feet from the cruiser. The angle between the water level and the submari
1 answer:
Answer:
Depth of the submarine = 668.09 feet
Step-by-step explanation:
From the figure attached,
In triangle ABC,
Distance between a navy cruiser and a nuclear submarine = 2500 feet
Angle between the water level and the submarine = 15.5°
By applying sine rule in the given triangle,
sin(∠BAC)° =
sin(15.5°) =
BC = 2500[sin(15.5)]
BC = 668.09 feet
Therefore, submarine is 668.09 feet deep in the sea.
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Answer: The ramp would be 15.5 feet long.
Step-by-step explanation: Please refer to the attached diagram for details.
Angle C shows the angle to be formed by the ramp from the ground, which is 15 degrees. Also, from the ground, it’s going to be four feet tall, which is line AB. The top of the ramp is point A, which makes line AC the entire length of the ramp. Since we have a reference angle (angle C) and two sides, the opposite and the hypotenuse, we shall apply the trigonometric ratio.
SinC = opposite/hypotenuse
Sin 15 = 4/b
By cross multiplication we now have
b = 4/Sin15
b = 4/0.2588
b = 15.4599
Approximately b = 15.5
Therefore the length of the ramp would be 15.5 feet
