The solution to this problem involves finding the length of line NL. Let's look at the given information:

MN and ML are two inclined lines that form angles of 30 and 60 degrees with the plane. MN and ML intersect at point N, while points M and L are located on the plane. In other words, we have a triangle MNL where angles MNL and MLN measure 30 and 60 degrees, respectively. We are trying to find the length of line NL.

First, let's use the law of sines to find the length of line NL:

NL/sin(30) = MN/sin(60)

We know that MN is equal to a, the distance between point M and the plane. We also know that the length of extension MA of line MNL is equal to a, as it forms a right angle with the plane. Therefore, we can rewrite the law of sines as:

NL/sin(30) = a/sin(60)

Next, we can rewrite sin(60) as cos(30). This gives us:

NL/sin(30) = a/cos(30) => NL = a/sqrt(3)

Therefore, the length of line NL is equal to a divided by the square root of 3, or a over root 3. This is the final answer to the problem.

Note that the distance 'a' should be provided in the same unit of measurement as the distance 'р'; otherwise, conversion will be needed using appropriate conversion formulas.

The side of a triangle is equal to 14, and the angles adjacent to it are 45 and 105 degrees. To find the side opposite to the angle in question, we can use the Law of Sines.

The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. In other words, a/Sin(A) = b/Sin(B) = c/Sin(C), where a, b, and c are the sides of the triangle and A, B, and C are the opposite angles.

Applying this to our problem, we can set up the following equation: 14/Sin(45°) = x/Sin(105°), where 14 is the known side, x is the unknown side, and 45° and 105° are the known adjacent angles.

Solving for x, we get x = 14 * Sin(105°)/Sin(45°) = 21.73. Therefore, the side opposite to the angle in question is approximately 21.73 units long.

Remember, always draw a diagram and label the sides and angles correctly when solving for unknowns in a triangle. It helps to visualize the problem and avoid confusion.