Maths Help????!!!!!!?

a tetrahedron is a polyhedron (solid in three dimensions with flat faces and straight edges) composed of four triangular faces wherein three of which meet at each vertex. A regular tetrahedron is one in which the four triangles are equilateral, meaning having equal sides. An equilateral triangle has also equal angles of 60 degrees.

therefore the angle between any two face is always equal to 60 degrees regardless of the length size.

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the northward line, the line of sight to the lighthouse at 8 pm and the line of sight to the lighthouse at 9 pm forms an obtuse triangle with one side equal to 18 km, the northward line.

At 9 pm, when the bearing is 40 degrees, the supplementary angle which is included in the triangle is (180 - 40) 140 degrees.

The bearing at 8 pm which is 25 degrees is one of the angles of the triangle.

The total of the three angles of a triangle should total 180 degrees, therefore,

the angle opposite the 18 km line is (180 - 140 - 25) 15 degrees,

which is the angle between the line of sight at the lighthouse.

we will use the law of sines which states

a / sin A = b / sin B = c / sin C

let n = distance of the lighthouse from the ship at 9.00 pm

18 / sin 15 = n / sin 25

n = 18 sin 25 / sin 15

....= 29.4 kms.

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at 9 pm, the line of sight to the lighthouse (which is 29.4 kms and a bearing of 40 degrees), the northward line and the line when the ship is due west of the lighthouse forms a right triangle with the line of sight at 9pm as the hypotenuse.

let w = distance from the 9pm point to the point when the ship will be due west of the lighthouse

cos 40 = w / 29.4

w = 29.4 cos 40

....= 22.515 kms.

since we know that the ship is travelling at 18 km / h

we use this formula to compute for the time when the ship will be due west of the lighthouse

speed = distance / time

time = distance / speed

........= 22.515 / 18

........= 1.25 hours

........= 75 minutes