A tower is broken at a point P above the ground. The top of the tower makes an angle `60^(@)` with the ground at Q. From another point R on the opposite side of Q angle of elevation of point P is `30^(@).` If `QR=180` meter, what is the total height (in meter) of the tower ?

Options

A. 90
B. `45sqrt3`
C. `45 (sqrt3 +1)`
D. `45 (sqrt3+2)`

Answer

To solve the problem, we need to find the total height of the tower using the information provided about the angles of elevation and the distance between points Q and R.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let the height of the tower be represented as \( H \).
- Let the height of the broken part of the tower (from point P to the ground) be \( PK \).
- Let the distance from point Q to the point directly below P (point K) be \( KQ \).
- The distance from point R to point Q is given as \( QR = 180 \) meters.

2. **Using Triangle PQK:**
- In triangle \( PQK \), we know:
- Angle \( Q = 60^\circ \)
- We can use the tangent function:
\[ \tan(60^\circ) = \frac{PK}{KQ} \]
- Since \( \tan(60^\circ) = \sqrt{3} \), we have:
\[ \sqrt{3} = \frac{PK}{KQ} \implies PK = \sqrt{3} \cdot KQ \]

3. **Using Triangle PRK:**
- In triangle \( PRK \), we know:
- Angle \( R = 30^\circ \)
- We can use the tangent function:
\[ \tan(30^\circ) = \frac{PK}{RK} \]
- Since \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{PK}{RK} \implies PK = \frac{RK}{\sqrt{3}} \]

4. **Relate RK and KQ:**
- From the information given:
\[ QR = RK + KQ = 180 \text{ meters} \]
- Let \( KQ = x \) meters, then \( RK = 180 - x \) meters.

5. **Substituting into the Equations:**
- From \( PK = \sqrt{3} \cdot KQ \):
\[ PK = \sqrt{3}x \]
- From \( PK = \frac{RK}{\sqrt{3}} \):
\[ PK = \frac{180 - x}{\sqrt{3}} \]

6. **Equating the Two Expressions for PK:**
\[ \sqrt{3}x = \frac{180 - x}{\sqrt{3}} \]
- Cross-multiplying gives:
\[ 3x^2 = 180 - x \]
- Rearranging:
\[ 3x^2 + x - 180 = 0 \]

7. **Using the Quadratic Formula:**
- The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 3, b = 1, c = -180 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-180)}}{2 \cdot 3} \]
\[ x = \frac{-1 \pm \sqrt{1 + 2160}}{6} = \frac{-1 \pm \sqrt{2161}}{6} \]

8. **Calculating the Value of x:**
- We only take the positive root:
\[ x \approx \frac{-1 + 46.5}{6} \approx 7.58 \text{ meters} \]

9. **Finding PK:**
- Now substitute \( x \) back to find \( PK \):
\[ PK = \sqrt{3} \cdot 7.58 \approx 13.1 \text{ meters} \]

10. **Finding the Total Height of the Tower:**
- The total height \( H = PK + KQ = PK + x \):
\[ H = 13.1 + 7.58 \approx 20.68 \text{ meters} \]

### Final Calculation:
- The total height of the tower is approximately \( 20.68 \) meters.