Let PM be the perpendicular from the point `P(1, 2, 3)` to XY-plane. If OP makes an angle `theta` with the positive direction of the Z-axies and OM makes an angle `Phi` with the positive direction of X-axis, where O is the origin, and `theta and Phi` are acute angles , then

Answer

To solve the problem step by step, we will analyze the situation involving the point \( P(1, 2, 3) \), the angles \( \theta \) and \( \phi \), and the relationships between them.

### Step 1: Understanding the Geometry
The point \( P(1, 2, 3) \) has coordinates in a three-dimensional space. The perpendicular from \( P \) to the XY-plane will intersect the XY-plane at the point \( M(1, 2, 0) \).

### Step 2: Establishing the Relationships
Let \( O \) be the origin \( (0, 0, 0) \). The vector \( OP \) can be represented as:
\[ OP = (1, 2, 3) \]
The length of this vector \( r \) is given by:
\[ r = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Step 3: Using Spherical Coordinates
In spherical coordinates, the coordinates can be expressed in terms of \( r \), \( \theta \), and \( \phi \):
- \( x = r \sin \theta \cos \phi \)
- \( y = r \sin \theta \sin \phi \)
- \( z = r \cos \theta \)

### Step 4: Setting Up the Equations
From the coordinates of point \( P \):
1. \( 1 = r \sin \theta \cos \phi \) (Equation 1)
2. \( 2 = r \sin \theta \sin \phi \) (Equation 2)
3. \( 3 = r \cos \theta \) (Equation 3)

### Step 5: Substitute \( r \)
We already found \( r = \sqrt{14} \). Substituting \( r \) into the equations gives:
1. \( 1 = \sqrt{14} \sin \theta \cos \phi \)
2. \( 2 = \sqrt{14} \sin \theta \sin \phi \)
3. \( 3 = \sqrt{14} \cos \theta \)

### Step 6: Simplifying the Equations
From Equation 3:
\[ \cos \theta = \frac{3}{\sqrt{14}} \quad \Rightarrow \quad \theta = \cos^{-1}\left(\frac{3}{\sqrt{14}}\right) \]

From Equation 1:
\[ \sin \theta \cos \phi = \frac{1}{\sqrt{14}} \quad \Rightarrow \quad \sin \theta = \sqrt{1 - \left(\frac{3}{\sqrt{14}}\right)^2} = \sqrt{\frac{5}{14}} \]
Substituting \( \sin \theta \) into Equation 1 gives:
\[ \sqrt{\frac{5}{14}} \cos \phi = \frac{1}{\sqrt{14}} \quad \Rightarrow \quad \cos \phi = \frac{1}{\sqrt{5}} \]

From Equation 2:
\[ \sin \theta \sin \phi = \frac{2}{\sqrt{14}} \quad \Rightarrow \quad \sin \phi = \frac{2}{\sqrt{14 \cdot \frac{5}{14}}} = \frac{2}{\sqrt{5}} \]

### Step 7: Finding the Angles
Now we can find \( \phi \):
\[ \tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = 2 \quad \Rightarrow \quad \phi = \tan^{-1}(2) \]

### Final Result
Thus, we have found the angles \( \theta \) and \( \phi \) in terms of their trigonometric relationships.