Question

Emma is on a 50 m high bridge and sees two boats anchored below. From her position, boat A has a bearing of 230° and boat B has a bearing of 120°. Emma estimates the angles of depression to be about 38° for boat A and 35° for boat B. How far apart are the boats to the nearest meter?

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Jayne

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In this diagram, point A represents Emma's position on the bridge, point B represents boat A, and point C represents boat B. We know that the height of the bridge is 50 meters, and the angles of depression to boats A and B are 38° and 35°, respectively. We also know that the bearings of boats A and B are 230° and 120°, respectively.
We can use the tangent function to find the distances from Emma to boats A and B. For boat A, we have:
tan(38°) = 50 / d_A
Solving for d_A, we get:
d_A = 50 / tan(38°) = 64.1 meters
Similarly, for boat B, we have:
tan(35°) = 50 / d_B
Solving for d_B, we get:
d_B = 50 / tan(35°) = 71.4 meters
Now that we know the distances from Emma to boats A and B, we can use the Law of Cosines to find the distance between the two boats. The Law of Cosines states that:
c^2 = a^2 + b^2 - 2ab * cos(C)
where c is the distance between the two points, a and b are the distances to the two points from a third point, and C is the angle between the lines connecting the third point to the two other points. In this case, we have:
c = d_A + d_B
a = 50 meters
b = 50 meters
C = 110°
Substituting these values into the Law of Cosines, we get:
(d_A + d_B)^2 = 50^2 + 50^2 - 2 * 50 * 50 * cos(110°)
Solving for c, we get:
c = sqrt((d_A + d_B)^2 - 50^2 - 50^2 + 2 * 50 * 50 * cos(110°))
Plugging in the values for d_A and d_B, we get:
c = sqrt((64.1 + 71.4)^2 - 50^2 - 50^2 + 2 * 50 * 50 * cos(110°))
Evaluating this expression, we get:
c = 111.4 meters
Therefore, the distance between the two boats is approximately 111 meters to the nearest meter.

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