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motikmotik
3 years ago
15

A lighthouse keeper views a small boat at an angle of depression of 62 degrees. The boat is 446 feet away from the base of the l

ighthouse. How tall is the lighthouse?

Mathematics
1 answer:
Genrish500 [490]3 years ago
4 0

Answer: the height of the lighthouse is 838.8 feet

Step-by-step explanation:

The right angle triangle ABC illustrating the scenario is shown in the attached photo.

The angle of depression and angle A are alternate angles, hence, they are the equal.

The height, h of the lighthouse represents the opposite side of the right angle triangle. The distance of the boat from the foot of the lighthouse represents the adjacent side of the right angle triangle.

To determine h, we would apply

the tangent trigonometric ratio.

Tan θ, = opposite side/adjacent side. Therefore,

Tan 62 = h/446

h = 446tan62 = 446 × 1.8807

h = 838.8 feet to the nearest tenth.

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Step-by-step explanation:

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2 years ago
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Solve the following proportion problem for x: 14/3x=20/x-5
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----  =  -----  cross multiply

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or

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I am very sorry because this is off-topic but is it normal to have 840/500 points? for ranking?
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2 years ago
A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line
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Answer:

Volume is 2000\pi\ m^{3}

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

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h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d              (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d                  (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

c = \frac{1}{5}

Therefore,

h(y) = \frac{1}{5}y + 5

Now, the Volume of the pool is given by:

V = \int h(y)dA

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A = r\theta

A = rdr\ d\theta

Thus

V = \int (\frac{1}{5}y + 5)dA

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta

V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta

V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta

V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}

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7 0
3 years ago
Cone B is the image of cone A after dilation by a scale factor of \frac{1}{2}
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An equation is an expression that shows the relationship between two or more numbers and variables.

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