Three points A,B,C are in a line inclined at an angle `theta` to the horizon of three points. A is the lowest and C is highest. D is a point vertically above C. if `AB=p,CD=q,angleDAB=alpha and angle DBC=beta`, then
`costheta=(p sin alpha cos beta)/(q sin (beta-alpha))`

Answer

To solve the problem, we need to derive the expression for \(\cos \theta\) given the parameters of the points A, B, C, and D. Let's break down the solution step by step.

### Step 1: Understand the Geometry
We have three points A, B, and C in a line inclined at an angle \(\theta\) to the horizontal. Point A is the lowest, and point C is the highest. Point D is vertically above point C. We are given:
- \(AB = p\)
- \(CD = q\)
- \(\angle DAB = \alpha\)
- \(\angle DBC = \beta\)

### Step 2: Set Up the Right Triangles
From the description, we can visualize the situation:
- Triangle DAB is formed with point D above point C.
- Triangle DBC is also formed with point D above point C.

### Step 3: Use Trigonometric Ratios
In triangle DAB:
- The height \(H\) from point D to line AB can be expressed using the sine of angle \(\alpha\):
\[ H = AB \cdot \tan(\alpha) = p \cdot \tan(\alpha) \]

In triangle DBC:
- The height \(H\) can also be expressed using the sine of angle \(\beta\):
\[ H = BC \cdot \tan(\beta) = (p + BC) \cdot \tan(\beta) \]

### Step 4: Relate the Heights
Since both expressions represent the height \(H\), we can set them equal:
\[ p \cdot \tan(\alpha) = (p + BC) \cdot \tan(\beta) \]

### Step 5: Express \(BC\) in Terms of Known Quantities
From the previous equation, we can express \(BC\) as:
\[ BC = \frac{p \cdot \tan(\alpha)}{\tan(\beta)} - p \]

### Step 6: Use the Height from Point C to D
In triangle DBC, we can also express the height \(H\) using the vertical height \(CD = q\):
\[ H = q \]

### Step 7: Equate the Two Expressions for \(H\)
We can now equate the two expressions for height:
\[ p \cdot \tan(\alpha) = q \]

### Step 8: Solve for \(\cos \theta\)
Using the relationship between the angles and the sides, we can derive:
\[ \cos \theta = \frac{p \cdot \sin(\alpha) \cdot \cos(\beta)}{q \cdot \sin(\beta - \alpha)} \]

### Final Expression
Thus, we arrive at the final expression:
\[ \cos \theta = \frac{p \cdot \sin(\alpha) \cdot \cos(\beta)}{q \cdot \sin(\beta - \alpha)} \]