A vertical pole PS has two marks at Q and R such that the portions PQ, PR and PS subtend angles `alpha,beta,gamma` at a point on the ground distant x from the pole. If PQ=a, PR=b,PS=c and `alpha+beta+gamma=180^(@)`, then `x^(2)=`

Options

A. `(a)/(a+b+c)`
B. `(b)/(a+b+c)`
C. `(c)/(a+b+c)`
D. `(abc)/(a+b+c)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem and draw a diagram
We have a vertical pole PS with two marks Q and R. The angles subtended by the segments PQ, PR, and PS at a point A on the ground are given as α, β, and γ respectively. The distances are PQ = a, PR = b, and PS = c. We need to find the value of x², where x is the distance from point A to the base of the pole.

**Hint:** Draw a vertical line for the pole and mark points P, Q, R, and S. Label the angles and distances accordingly.

### Step 2: Set up the relationships using trigonometry
From triangle APQ:
\[ \tan(\alpha) = \frac{PQ}{x} = \frac{a}{x} \]
Thus, we can express:
\[ a = x \tan(\alpha) \]

From triangle APR:
\[ \tan(\beta) = \frac{PR}{x} = \frac{b}{x} \]
Thus, we can express:
\[ b = x \tan(\beta) \]

From triangle ASP:
\[ \tan(\gamma) = \frac{PS}{x} = \frac{c}{x} \]
Thus, we can express:
\[ c = x \tan(\gamma) \]

**Hint:** Use the definition of tangent in right triangles to relate the lengths and angles.

### Step 3: Combine the equations
Now, we have:
1. \( a = x \tan(\alpha) \)
2. \( b = x \tan(\beta) \)
3. \( c = x \tan(\gamma) \)

### Step 4: Use the identity for angles
Since \( \alpha + \beta + \gamma = 180^\circ \), we can use the tangent addition formula:
\[ \tan(\alpha + \beta + \gamma) = \tan(180^\circ) = 0 \]
This leads to the identity:
\[ \tan(\alpha) + \tan(\beta) + \tan(\gamma) = \tan(\alpha) \tan(\beta) \tan(\gamma) \]

### Step 5: Substitute the expressions
Substituting the expressions we derived:
\[ \frac{a}{x} + \frac{b}{x} + \frac{c}{x} = \frac{abc}{x^2} \]

Multiplying through by \( x \) gives:
\[ a + b + c = \frac{abc}{x} \]

### Step 6: Solve for x
Rearranging gives:
\[ x = \frac{abc}{a + b + c} \]

### Step 7: Find x²
Now, squaring both sides:
\[ x^2 = \left(\frac{abc}{a + b + c}\right)^2 \]

**Final Answer:**
\[ x^2 = \frac{a^2 b^2 c^2}{(a + b + c)^2} \]