The angle of elevation of a cliff from a fixed point A is `45^@` . After going up a distance of 600 metres towards the top of the cliff at an inclination of `30^@`, it is found that the angle of elevation is `60^@` . Find the height of the cliff. (Use `sqrt3 = 1.732`)

Options

A. 817.8 m
B. 818.5 m
C. 820.5 m
D. 819.6 m

Answer

To solve the problem step by step, we will break it down into manageable parts using trigonometric principles.

### Step 1: Understand the Problem
We have a cliff and two points: point A (the initial observation point) and point P (after moving towards the cliff). We know the angles of elevation from these points and the distance moved.

### Step 2: Set Up the Diagram
1. Let the height of the cliff be \( H \).
2. From point A, the angle of elevation to the top of the cliff is \( 45^\circ \).
3. After moving 600 meters towards the cliff at an inclination of \( 30^\circ \), the angle of elevation from point P becomes \( 60^\circ \).

### Step 3: Use Triangle APM
In triangle APM:
- Angle \( APM = 30^\circ \)
- Distance \( AP = 600 \) meters (the distance moved towards the cliff).

Using the sine function:
\[ \sin(30^\circ) = \frac{PM}{AP} \]
\[ \sin(30^\circ) = \frac{1}{2} \implies \frac{PM}{600} = \frac{1}{2} \implies PM = 300 \text{ meters} \]

### Step 4: Find Base AM
Using the cosine function:
\[ \cos(30^\circ) = \frac{AM}{AP} \]
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \implies \frac{AM}{600} = \frac{\sqrt{3}}{2} \implies AM = 300\sqrt{3} \text{ meters} \]

### Step 5: Use Triangle BPN
In triangle BPN:
- Angle \( BPN = 60^\circ \)
- The height from point P to the top of the cliff is \( H - PM = H - 300 \).

Using the tangent function:
\[ \tan(60^\circ) = \frac{BN}{PN} \]
\[ \tan(60^\circ) = \sqrt{3} \implies \frac{H - 300}{PN} = \sqrt{3} \]
Thus, \( PN = \frac{H - 300}{\sqrt{3}} \).

### Step 6: Find AC
Now, we find the total distance \( AC \):
\[ AC = AM + MC = AM + PN \]
Substituting the values:
\[ AC = 300\sqrt{3} + \frac{H - 300}{\sqrt{3}} \]
Finding a common denominator:
\[ AC = \frac{300\sqrt{3} \cdot \sqrt{3} + H - 300}{\sqrt{3}} = \frac{900 + H - 300}{\sqrt{3}} = \frac{H + 600}{\sqrt{3}} \]

### Step 7: Use Triangle ABC
In triangle ABC:
\[ \tan(45^\circ) = \frac{BC}{AC} \]
Since \( \tan(45^\circ) = 1 \):
\[ 1 = \frac{H}{\frac{H + 600}{\sqrt{3}}} \]
Cross-multiplying gives:
\[ H + 600 = \sqrt{3}H \]
Rearranging:
\[ 600 = \sqrt{3}H - H = (\sqrt{3} - 1)H \]
Thus:
\[ H = \frac{600}{\sqrt{3} - 1} \]

### Step 8: Rationalize the Denominator
To simplify:
\[ H = \frac{600(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{600(\sqrt{3} + 1)}{3 - 1} = 300(\sqrt{3} + 1) \]
Substituting \( \sqrt{3} \approx 1.732 \):
\[ H = 300(1.732 + 1) = 300(2.732) \approx 819.6 \text{ meters} \]

### Final Answer
The height of the cliff is approximately \( 819.6 \) meters.

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