Help 15 points!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Svetllana [295]2 years ago

3 0

Answer:

see explanation

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan A = $\frac{opposite}{adjacent}$ = $\frac{BC}{AC}$ = $\frac{7}{8}$ , then

∠ A = $tan^{-1}$ ( $\frac{7}{8}$ ) ≈ 41° ( to the nearest degree )

The sum of the angles in the triangle = 180° , then

∠ B + 41° + 90° = 180°

∠ B + 131° = 180° ( subtract 131° from both sides )

∠ B = 49°

Using Pythagoras' identity in the right triangle

AB² = BC² + AC² = 7² + 8² = 49 + 64 = 113 ( take square root of both sides )

AB = $\sqrt{113}$ ≈ 10.6 ( to the nearest tenth )

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The depth of the water at the end of a pier changes periodically along with the movement of tides. On a particular day, low tide

Valentin [98]

Draw a diagram to illustrate the problem as shown in the figure below.

The minimum depth of 2.5 m occurs at 12:00 am and at 12:30 pm.
Therefore the period i0s T= 12.5 hours.
The maximum depth of 5.5 m occurs at 6:15 am and at 6:45 pm. Therefore the period of T = 12.5 hours is confirmed.

The double amplitude is 5.5 - 2.5 = 3 m, therefore the amplitude is a = 1.5 m.
The mean depth is k = (2.5 + 5.5)/2 = 4.0 m

The model for tide depth is
$d=-1.5cos( \frac{2 \pi t}{12.5})+4$

That is,
d = -1.5 cos(0.5027t) + 4
where
d = depth, m
t = time, hours

A plot of the function confirms that the model is correct.