Understanding the Problem: Flagstaff on a Pillar and Angles

This problem involves calculating the tangent of an angle subtended by a pillar at a point on the ground, given information about a flagstaff standing on top of the pillar and the angle it subtends at the same point. We need to use principles of trigonometry, specifically the tangent function and the tangent subtraction formula, along with basic geometry.

Setting Up the Geometry

Let's visualize the scenario:

• The pillar has a height of 10 m.
• The flagstaff on top of the pillar is 20 m long.
• The total height from the ground to the top of the flagstaff is the height of the pillar plus the length of the flagstaff, which is \(10 \, \text{m} + 20 \, \text{m} = 30 \, \text{m}\).
• Let P be a point on the ground.
• Let \(x\) be the horizontal distance from the base of the pillar to point P.
• Let \(\theta\) be the angle subtended by the pillar at point P. This is the angle of elevation to the top of the pillar from P.
• Let \(\phi\) be the angle subtended by the total height (pillar + flagstaff) at point P. This is the angle of elevation to the top of the flagstaff from P.

From the right triangle formed by the pillar, the ground, and the line of sight to the top of the pillar, we have:

\( \tan \theta = \frac{\text{Height of pillar}}{\text{Distance } x} = \frac{10}{x} \)

From the right triangle formed by the total height, the ground, and the line of sight to the top of the flagstaff, we have:

\( \tan \phi = \frac{\text{Total height}}{\text{Distance } x} = \frac{30}{x} \)

The angle subtended by the flagstaff itself at point P is the difference between the angle subtended by the total height and the angle subtended by the pillar. This angle is given as \(\tan^{-1}(0.5)\).

So, we have:

\( \phi - \theta = \tan^{-1}(0.5) \)

Taking the tangent of both sides:

\( \tan(\phi - \theta) = \tan(\tan^{-1}(0.5)) = 0.5 = \frac{1}{2} \)

Applying the Tangent Subtraction Formula

We will use the tangent subtraction formula, which states:

\( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)

In our case, \(A = \phi\) and \(B = \theta\). Substituting the values of \(\tan \phi\) and \(\tan \theta\) we found:

\( \tan(\phi - \theta) = \frac{\frac{30}{x} - \frac{10}{x}}{1 + \left(\frac{30}{x}\right) \left(\frac{10}{x}\right)} \)

We know that \(\tan(\phi - \theta) = \frac{1}{2}\). So, the equation becomes:

\( \frac{1}{2} = \frac{\frac{20}{x}}{1 + \frac{300}{x^2}} \)

Solving for the Distance x

Let's simplify the right side of the equation:

\( \frac{1}{2} = \frac{\frac{20}{x}}{\frac{x^2 + 300}{x^2}} \)

\( \frac{1}{2} = \frac{20}{x} \cdot \frac{x^2}{x^2 + 300} \)

\( \frac{1}{2} = \frac{20x}{x^2 + 300} \)

Now, cross-multiply:

\( 1 \cdot (x^2 + 300) = 2 \cdot (20x) \)

\( x^2 + 300 = 40x \)

Rearrange into a quadratic equation:

\( x^2 - 40x + 300 = 0 \)

We can solve this quadratic equation for \(x\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-40\), and \(c=300\).

\( x = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(1)(300)}}{2(1)} \)

\( x = \frac{40 \pm \sqrt{1600 - 1200}}{2} \)

\( x = \frac{40 \pm \sqrt{400}}{2} \)

\( x = \frac{40 \pm 20}{2} \)

This gives two possible values for \(x\):

• \( x_1 = \frac{40 + 20}{2} = \frac{60}{2} = 30 \)
• \( x_2 = \frac{40 - 20}{2} = \frac{20}{2} = 10 \)

Both positive values for \(x\) are physically possible distances.

Finding Possible Values of tan \(\theta\)

We defined \(\tan \theta = \frac{10}{x}\). We will now calculate \(\tan \theta\) for each possible value of \(x\).

• If \(x = 30\): \( \tan \theta = \frac{10}{30} = \frac{1}{3} \)
• If \(x = 10\): \( \tan \theta = \frac{10}{10} = 1 \)

The possible values of \(\tan \theta\) are \(\frac{1}{3}\) and \(1\).

Comparing with Options

Let's compare our possible values for \(\tan \theta\) with the given options:

• Option 1: \( \frac{3}{4} \)
• Option 2: \( \frac{2}{3} \)
• Option 3: \( \frac{1}{3} \)
• Option 4: \( \frac{1}{4} \)

The value \(\frac{1}{3}\) is one of the possible values we calculated and is present in the options.

Conclusion on Possible tan \(\theta\) Value

Based on our calculations, a possible value for \(\tan \theta\) is \(\frac{1}{3}\), which corresponds to one of the provided options.

Revision Table: Key Concepts in Trigonometry and Angles

Additional Information: Solving Trigonometry Problems

When tackling trigonometry problems involving heights and distances, especially with multiple objects or angles:

• Always draw a clear diagram. This helps visualize the right triangles involved and the angles of interest.
• Define variables for unknown quantities like distances or angles.
• Use the SOH CAH TOA mnemonic to relate sides and angles using sine, cosine, or tangent. For angles of elevation/depression, tangent is often useful when relating vertical height to horizontal distance.
• Look for relationships between angles. For instance, an angle subtended by an object positioned above another is the difference between two angles of elevation.
• Apply relevant trigonometric identities or formulas, such as the tangent addition or subtraction formulas.
• The problem may lead to solving algebraic equations, including linear or quadratic equations. Be prepared to solve these accurately.
• Always check if the solutions obtained are physically realistic (e.g., distances must be positive).