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Musya8 [376]
3 years ago
9

A series of three? separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers. Each

of the three tunnels is shaped like a? half-cylinder with a radius of 5 meters? (the height of the? tunnel) and a length of 25 kilometers? (the length of the? tunnel). How much earth? (volume) was removed to build the three? tunnels
Mathematics
1 answer:
Nadya [2.5K]3 years ago
4 0

Answer:

312.5\pi \text{ km}^3\approx 981.75\text{ km}^3

Step-by-step explanation:

We have been given that a series of 3 separate, adjacent tunnels is constructed through a mountain. Its length is approximately 25 kilometers.

Each of the three tunnels is shaped like a half-cylinder with a radius of 5 meters.

Since we know that volume of a semicircular or a half cylinder is half the volume of a circular cylinder.

\text{Volume of a semicircular cylinder}=\frac{\pi r^2h}{2}, where,

r = Radius of cylinder,

h = height of the cylinder.

Upon substituting our given values in volume formula we will get,

\text{Volume of a semicircular cylinder}=\frac{\pi (5\text{ km})^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*25\text{ km}^2*25\text{ km}}{2}

\text{Volume of a semicircular cylinder}=\frac{\pi*625\text{ km}^3}{2}

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

\text{Volume of a semicircular cylinder}=\pi*312.5\text{ km}^3

\text{Volume of a semicircular cylinder}=981.74770\text{ km}^3

Therefore, the volume of earth removed to build the three tunnels is 312.5\pi \text{ km}^3\approx 981.75\text{ km}^3.


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The distance between point on the ground from the top of the building is 396 meter, if the building is 280 m high and The angle of depression from the top of a building to a point  on the ground is 45 degrees.

Step-by-step explanation:

The given is,

                 The angle of depression from the top of a building to a point  on the ground is 45 degrees.

                 Height of the building is 280 meter.

Step: 1

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                              90° = 45° + 45°    

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                 Trignometric ratio,

                            sin ∅ = \frac{Opp}{Hyp}....................(1)

                For the above ratio take the bottom angle, that is angle of depression from the top of a building to a point  on the ground is 45 degrees.

                 Where, Opp side = 280 meters

                               Hyp side = x

                                           ∅ = 45°

                 Equation (1) becomes,

                                  sin 45° = \frac{280}{x}

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                                            x =  \frac{280}{0.70710678}

                                            x = 395.979

               Distance between point on the ground from the top of the building,  x ≅ 396 meter                                                

                Trignometric ratio,

                                     cos ∅ = \frac{Adj}{Hyp}

                                   Cos 45 = \frac{Adj}{396}

                                         Adj = (0.70710678)(396)

              Bottom length, Adj = 280 meter

Result:

           The distance between point on the ground from the top of the building is 396 meter.

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