Question

From the observation deck of a seaside building 200m high, Jagan sees two

fishing boats. The angle of depression to the nearest boat is 60° while for
the boat farther away the angle is 45°
(a) How far out to the sea is the nearest boat?
(b) How far apart are the two boats?

Answer 1

Explanation:

Let's denote the distance from the observation deck to the nearest boat as x, and the distance from the observation deck to the farther boat as y. We can use trigonometry to solve for x and y.

(a) To find x, we can use the tangent function:

tan(60°) = x/200

Solving for x, we get:

x = 200 tan(60°)

x ≈ 346.4 meters

Therefore, the nearest boat is about 346.4 meters away from the observation deck.

(b) To find y, we can use the tangent function again:

tan(45°) = y/200

Solving for y, we get:

y = 200 tan(45°)

y ≈ 200 meters

Therefore, the farther boat is about 200 meters away from the observation deck.

To find the distance between the two boats, we can simply subtract x from y:

y - x ≈ 200 - 346.4 ≈ -146.4 meters

This result is negative, which means that the two boats are actually closer than the observation deck. This could be due to a few reasons, such as the boats being located behind a cliff or a harbor wall. Alternatively, there could be an error in the measurements or calculations.

Answer 2

(a) 115.47 meters

(b) 261.5 meters

Explanation:

Let's label the points as follows: the top of the seaside building is point A, the location of the nearest boat is point B, and the location of the farther boat is point C. We know that AB = 200 m (the height of the building) and that angle BAD = 60° and angle CAD = 45°. We want to find the distance BC (part a) and the distance AC (part b).

(a) To find BC, we can use trigonometry. Let x be the distance from point B to the foot of the perpendicular dropped from point A to the sea (point D). Then, we have:

tan 60° = AB/BD

tan 60° = 200/x

x = 200/tan 60°

x = 200/√3

x ≈ 115.47

So the distance from the building to the nearest boat (BC) is approximately 115.47 meters.

(b) To find AC, we can use the fact that triangle ABC is a right triangle, with angle ABC = 180° - 60° - 45° = 75°. Then we have:

sin 75° = BC/AC

AC = BC/sin 75°

AC ≈ 261.5

So the distance between the two boats is approximately 261.5 meters.