Heights & Distances Questions and Answers

Answer: A

Both aircraft are on the same vertical line, so they share one horizontal distance \( d \) from the observer.

For the upper plane at 900 m, elevation 60°:

\[ \tan 60^{\circ} = \frac{900}{d} \quad\Longrightarrow\quad d = \frac{900}{\sqrt{3}} = 300\sqrt{3} = 519.6 \text{ m} \]

For the lower plane, elevation 45°:

\[ h = d\tan 45^{\circ} = 519.6 \text{ m} \]

\[ \text{difference} = 900 - 519.6 = 380.4 \approx 381 \text{ m} \]

The steeper 60° sighting must belong to the higher aircraft, which fixes which angle goes with which plane.

Answer: A

Let the tower be \( h \) m. The shadow shortens as the sun climbs.

\[ s_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}, \qquad s_{60} = \frac{h}{\tan 60^{\circ}} = \frac{h}{\sqrt{3}} \]

\[ h\sqrt{3} - \frac{h}{\sqrt{3}} = 70 \]

\[ \frac{3h - h}{\sqrt{3}} = 70 \quad\Longrightarrow\quad \frac{2h}{\sqrt{3}} = 70 \]

\[ h = 35\sqrt{3} = 35 \times 1.732 = 60.62 \approx 60.6 \text{ m} \]

70 m is the change in the shadow, not a shadow length - using it directly as one of the two lengths gives 140 m or 35 m, which is why both appear as distractors.

Answer: B

The angle of elevation \( \theta \) satisfies

\[ \tan\theta = \frac{\text{height of tree}}{\text{length of shadow}} \]

Here the shadow is \( \sqrt{3} \) times the height, so with height \( h \):

\[ \tan\theta = \frac{h}{\sqrt{3}\,h} = \frac{1}{\sqrt{3}} \]

\[ \theta = 30^{\circ} \]

The tree's actual height cancels, so it is never needed. Inverting the ratio gives \( \tan\theta = \sqrt{3} \) and hence 60° - the mirror-image case, where the shadow is shorter than the tree.

Answer: D

The tower, the ground and the line of sight form a right triangle with the 30° angle at \( P \).

\[ \tan 30^{\circ} = \frac{\text{height}}{\text{distance}} \quad\Longrightarrow\quad \text{distance} = \frac{\text{height}}{\tan 30^{\circ}} \]

\[ d = \frac{100}{1/\sqrt{3}} = 100\sqrt{3} \]

\[ d = 100 \times 1.732 = 173.2 \text{ m} \]

A 30° elevation always puts the observer \( \sqrt{3} \) times the height away. Dividing by \( \sqrt{3} \) instead of multiplying gives about 58 m, which is the classic sign error.

Answer: A

The ladder is the hypotenuse and the ground gap is the side adjacent to the 60° angle, so cosine is the ratio to use.

\[ \cos 60^{\circ} = \frac{\text{distance from wall}}{\text{ladder length}} \]

\[ \frac{1}{2} = \frac{4.6}{L} \quad\Longrightarrow\quad L = 2 \times 4.6 = 9.2 \text{ m} \]

Since \( \cos 60^{\circ} = \tfrac{1}{2} \), the ladder is always exactly twice the foot distance here. Reaching for \( \tan \) or \( \sin \) instead gives \( 4.6\sqrt{3} \) - the wall height, not the ladder.

Answer: B

The lighthouse foot lies between the ships, so the total distance is the sum of the two horizontal distances.

\[ d = \frac{h}{\tan 30^{\circ}} + \frac{h}{\tan 45^{\circ}} \]

With \( h = 100 \), \( \tan 30^{\circ} = \tfrac{1}{\sqrt{3}} \) and \( \tan 45^{\circ} = 1 \):

\[ d = 100\sqrt{3} + 100 = 173.2 + 100 = 273.2 \approx 273 \text{ m} \]

Because the ships are on opposite sides the distances add. Had they been on the same side the answer would be the difference, 73 m. Choosing 173 m means only the \( 30^{\circ} \) ship was counted.

Answer: C

The tower, the ground and the line of sight form a right triangle with the 30° angle at \( P \).

\[ \tan 30^{\circ} = \frac{\text{height}}{\text{distance}} \quad\Longrightarrow\quad \text{distance} = \frac{\text{height}}{\tan 30^{\circ}} \]

\[ d = \frac{200}{1/\sqrt{3}} = 200\sqrt{3} \]

\[ d = 200 \times 1.732 = 346.4 \text{ m} \]

A 30° elevation always puts the observer \( \sqrt{3} \) times the height away. Dividing by \( \sqrt{3} \) instead of multiplying gives about 115 m, which is the classic sign error.

Answer: D

Height and shadow are the two legs of the right triangle formed by the sun's ray, and the angle of elevation \( \theta \) sits at the tip of the shadow.

\[ \tan\theta = \frac{\text{height of tree}}{\text{length of shadow}} \]

The two are equal, so the ratio is 1:

\[ \tan\theta = 1 \quad\Longrightarrow\quad \theta = 45^{\circ} \]

No length is needed - only the ratio matters, which is why the question can omit the tree's actual height. At \( 30^{\circ} \) the shadow would be \( \sqrt{3} \) times the height; at \( 60^{\circ} \), \( \tfrac{1}{\sqrt{3}} \) of it.

Answer: B

The angle is measured from the observer's eye, 2 m above the ground, so the trigonometry gives only the part of the tower above eye level.

\[ \tan 30^{\circ} = \frac{x}{10\sqrt{3}} \]

\[ x = 10\sqrt{3} \times \frac{1}{\sqrt{3}} = 10 \text{ m} \]

Adding the observer's own height:

\[ \text{tower} = 10 + 2 = 12 \text{ m} \]

Forgetting to add the 2 m gives 10 m, which is offered as the first distractor. The \( 10\sqrt{3} \) is chosen so the radical cancels exactly.

Answer: B

Let the tower be \( T \) m high and \( d \) m away. Two sightings:

\[ \tan b = \frac{T}{d} = \frac{4}{5}, \qquad \tan a = \frac{T - 230}{d} = \frac{5}{12} \]

Subtracting eliminates \( T \):

\[ \frac{230}{d} = \frac{4}{5} - \frac{5}{12} = \frac{48 - 25}{60} = \frac{23}{60} \]

\[ d = 230 \times \frac{60}{23} = 600 \text{ m} \]

\[ T = \frac{4}{5} \times 600 = 480 \text{ m}, \qquad T - 230 = 250 \text{ m} \]

The line joining the two tops is the hypotenuse of a right triangle with legs \( d \) and \( T - 230 \):

\[ \sqrt{600^{2} + 250^{2}} = \sqrt{360000 + 62500} = \sqrt{422500} = 650 \text{ m} \]

600 m is the horizontal distance only; the question asks for the slant distance between the tops.