Calculating Flagstaff Height Using Ladder and Elevation Angle

This problem involves using trigonometry and geometry to find the height of a flagstaff given information about a ladder leaning against it and an angle of elevation.

Understanding the Geometry

Let's visualize the scenario and label the points:

• Let A be the foot of the ladder on the ground.
• Let C be the base of the vertical flagstaff on the ground.
• Let D be the top of the flagstaff.
• Let B be the point on the flagstaff where the ladder touches it.

We are given:

• The ladder length AB = 6 m.
• The ladder reaches a point B which is 6 m below the top D. So, BD = 6 m.
• The flagstaff is vertical, so the angle ∠ACD is 90°.
• The angle of elevation of the top of the flagstaff from the foot of the ladder is ∠CAD = 75°.

Let the total height of the flagstaff be H meters. Thus, CD = H.

The point B is on the flagstaff, so C, B, and D are collinear. Since BD = 6 m and CD = H, the height of point B from the base C is BC = CD - BD = H - 6.

Triangle ACD is a right-angled triangle at C. Triangle ACB is also a right-angled triangle at C.

Using Trigonometry in Triangle ACD

In the right-angled triangle ACD, we have the angle of elevation ∠CAD = 75° and the side CD = H. The distance from the foot of the ladder to the base of the flagstaff is AC.

We can use the tangent function:

\[ \tan(\angle CAD) = \frac{CD}{AC} \] \[ \tan(75^\circ) = \frac{H}{AC} \]

To find AC, we need the value of \(\tan(75^\circ)\). We can calculate this using the sum of angles formula for tangent:

\[ \tan(75^\circ) = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \]

We know that \(\tan 45^\circ = 1\) and \(\tan 30^\circ = \frac{1}{\sqrt{3}}\). Substituting these values:

\[ \tan(75^\circ) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \]

To rationalize the denominator, multiply the numerator and denominator by \((\sqrt{3} + 1)\):

\[ \tan(75^\circ) = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3})^2 + 2\sqrt{3} + 1}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \]

So, \(\tan(75^\circ) = 2 + \sqrt{3}\).

Now, from the equation \(\tan(75^\circ) = \frac{H}{AC}\), we get:

\[ AC = \frac{H}{\tan(75^\circ)} = \frac{H}{2 + \sqrt{3}} \]

Using Pythagorean Theorem in Triangle ACB

In the right-angled triangle ACB, the hypotenuse is the ladder length AB = 6 m. The sides are AC and BC. We know BC = H - 6.

Applying the Pythagorean theorem:

\[ AC^2 + BC^2 = AB^2 \] \[ AC^2 + (H - 6)^2 = 6^2 \] \[ AC^2 + (H - 6)^2 = 36 \]

Connecting the Two Triangles

We have two expressions involving AC: one from ▵ACD (\(AC = \frac{H}{2 + \sqrt{3}}\)) and one from ▵ACB (\(AC^2 + (H - 6)^2 = 36\)). Let's substitute the expression for AC into the second equation:

\[ \left(\frac{H}{2 + \sqrt{3}}\right)^2 + (H - 6)^2 = 36 \] \[ \frac{H^2}{(2 + \sqrt{3})^2} + (H - 6)^2 = 36 \] \[ \frac{H^2}{4 + 3 + 4\sqrt{3}} + (H - 6)^2 = 36 \] \[ \frac{H^2}{7 + 4\sqrt{3}} + (H - 6)^2 = 36 \]

This is a quadratic equation in terms of H. Solving this directly might be complex. Let's check the provided options, as they are in a specific form.

Checking the Options

Let's test the option \(H = (6 + 3\sqrt{3})\) m.

If \(H = 6 + 3\sqrt{3}\), then \(H - 6 = 3\sqrt{3}\).

Now calculate AC using \(AC = \frac{H}{2 + \sqrt{3}}\) with this value of H:

\[ AC = \frac{6 + 3\sqrt{3}}{2 + \sqrt{3}} \]

Rationalize the denominator:

\[ AC = \frac{6 + 3\sqrt{3}}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{(6)(2) - 6\sqrt{3} + (3\sqrt{3})(2) - (3\sqrt{3})(\sqrt{3})}{2^2 - (\sqrt{3})^2} \] \[ AC = \frac{12 - 6\sqrt{3} + 6\sqrt{3} - 3 \times 3}{4 - 3} = \frac{12 - 9}{1} = 3 \]

So, if \(H = 6 + 3\sqrt{3}\), then \(AC = 3\) and \(H - 6 = 3\sqrt{3}\).

Now substitute these values into the Pythagorean equation for ▵ACB: \(AC^2 + (H - 6)^2 = 36\).

\[ 3^2 + (3\sqrt{3})^2 = 9 + (9 \times 3) = 9 + 27 = 36 \]

This matches \(6^2\), which is the square of the ladder length AB. Thus, the height \(H = (6 + 3\sqrt{3})\) m satisfies the conditions of the problem.

Conclusion

Based on the calculations, the height of the flagstaff is \((6 + 3\sqrt{3})\) m.

Revision Table: Key Values

Additional Information: Understanding Special Angles

The angle 75° is not a standard angle like 0°, 30°, 45°, 60°, or 90°, but its trigonometric ratios can be calculated using the sum or difference of standard angles (e.g., 45° + 30° or 90° - 15°).

Key trigonometric identities used:

• Tangent addition formula: \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\).
• Pythagorean theorem in a right triangle: \(a^2 + b^2 = c^2\).

Knowing the values of trigonometric ratios for 30°, 45°, and 60° is essential for calculating ratios for angles like 15° or 75°.

• \(\sin 30^\circ = 1/2\), \(\cos 30^\circ = \sqrt{3}/2\), \(\tan 30^\circ = 1/\sqrt{3}\)
• \(\sin 45^\circ = 1/\sqrt{2}\), \(\cos 45^\circ = 1/\sqrt{2}\), \(\tan 45^\circ = 1\)
• \(\sin 60^\circ = \sqrt{3}/2\), \(\cos 60^\circ = 1/2\), \(\tan 60^\circ = \sqrt{3}\)

These values are fundamental in solving problems involving various angles in trigonometry and geometry.