Heights & Distances Questions and Answers

Answer: B

The ladder is the hypotenuse and the 3 m is adjacent to the \( 45^{\circ} \) angle, so cosine relates them.

\[ \cos 45^{\circ} = \frac{3}{L} \quad\Longrightarrow\quad \frac{1}{\sqrt{2}} = \frac{3}{L} \]

\[ L = 3\sqrt{2} = 3 \times 1.414 = 4.24 \approx 4.23 \text{ m} \]

At \( 45^{\circ} \) the wall height also equals 3 m, and the ladder is the diagonal of a 3 m square - hence the \( \sqrt{2} \).

Answer: C

The cable has no slack, so it is a straight hypotenuse of 200 m, and the balloon's height is the side opposite the \( 60^{\circ} \) inclination.

\[ \sin 60^{\circ} = \frac{h}{200} \]

\[ h = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3} = 173.2 \text{ m} \]

The cable length is a slant distance, so sine applies, not tangent. The height must come out below 200 m, which rules out anything larger on inspection.

Answer: C

Both ships lie on the same side of the helicopter, so the distance between them is the difference of their horizontal distances from the point directly below it.

\[ d_{60} = \frac{1200}{\tan 60^{\circ}} = \frac{1200}{\sqrt{3}} = 400\sqrt{3} = 692.8 \text{ m} \]

\[ d_{45} = \frac{1200}{\tan 45^{\circ}} = 1200 \text{ m} \]

\[ \text{separation} = 1200 - 400\sqrt{3} = 400\left(3 - \sqrt{3}\right) \]

\[ = 400 \times 1.268 = 507.2 \text{ m} \]

The steeper \( 60^{\circ} \) depression belongs to the nearer ship. Stopping at \( 400\sqrt{3} \) gives that ship's distance from the helicopter's base, not the gap between the two.

Answer: A

Let the tower be \( h \) m and the observation point \( d \) m away. The bottom of the flagstaff is the top of the tower.

\[ \tan 30^{\circ} = \frac{h}{d} \quad\Longrightarrow\quad d = h\sqrt{3} \]

\[ \tan 45^{\circ} = \frac{h+7}{d} \quad\Longrightarrow\quad h + 7 = d = h\sqrt{3} \]

\[ h(\sqrt{3} - 1) = 7 \quad\Longrightarrow\quad h = \frac{7}{\sqrt{3}-1} = \frac{7(\sqrt{3}+1)}{2} \]

\[ h = 3.5 \times 2.732 = 9.562 \text{ m} \]

Both sightings share the same \( d \), which is what allows one to be substituted into the other.

Answer: D

Let \( d \) be the horizontal distance between the building and the tower.

The tower's foot is 30 m below the observer, at a depression of \( 45^{\circ} \):

\[ d = \frac{30}{\tan 45^{\circ}} = 30 \text{ m} \]

The tower's top, at a depression of \( 30^{\circ} \), is

\[ d\tan 30^{\circ} = \frac{30}{\sqrt{3}} \text{ m} \]

below the building's top, so the tower's height is

\[ 30 - \frac{30}{\sqrt{3}} = 30\left(1 - \frac{1}{\sqrt{3}}\right) \approx 12.7 \text{ m} \]

The tower is shorter than the building, which is consistent with both angles being depressions rather than elevations.

Answer: B

Let the tower be \( H \) m and \( d \) m away, and call the common angle \( \theta \).

The depression of the tower's foot uses the cliff's own height:

\[ \tan\theta = \frac{25}{d} \]

The elevation of the tower's top uses the part above the cliff:

\[ \tan\theta = \frac{H - 25}{d} \]

The two are equal, so the numerators must match:

\[ H - 25 = 25 \quad\Longrightarrow\quad H = 50 \text{ m} \]

Whenever the elevation of a top equals the depression of a foot, the object is exactly twice the observer's height - the distance never enters the answer.

Answer: C

The shadow is the base and the pole is opposite the sun's \( 60^{\circ} \) altitude.

\[ \tan 60^{\circ} = \frac{h}{30} \]

\[ h = 30\sqrt{3} \approx 51.96 \text{ m} \]

With the sun above \( 45^{\circ} \), the pole is taller than its shadow - so any answer under 30 m is wrong by inspection. Dividing instead of multiplying gives \( 10\sqrt{3} \), the first distractor.

Answer: A

The building is opposite the \( 30^{\circ} \) altitude and the shadow is the base.

\[ \tan 30^{\circ} = \frac{50}{s} \quad\Longrightarrow\quad s = \frac{50}{\tan 30^{\circ}} \]

\[ s = 50\sqrt{3} \approx 86.6 \text{ m} \]

A low sun casts a long shadow, so the answer must exceed the 50 m height - a quick check that the division went the right way round.

Answer: C

Compute both shadow lengths for the 15 m pole and subtract.

\[ s_{30} = \frac{15}{\tan 30^{\circ}} = 15\sqrt{3} \text{ m} \]

\[ s_{60} = \frac{15}{\tan 60^{\circ}} = \frac{15}{\sqrt{3}} = 5\sqrt{3} \text{ m} \]

\[ \text{difference} = 15\sqrt{3} - 5\sqrt{3} = 10\sqrt{3} \approx 17.3 \text{ m} \]

Writing \( \tfrac{15}{\sqrt{3}} \) as \( 5\sqrt{3} \) puts both terms in the same surd, so they subtract directly.

Answer: D

Join the two tops. That line's vertical leg is the height difference and its horizontal leg is the gap between the poles.

\[ \text{height difference} = 80 - 62.5 = 17.5 \text{ m} \]

\[ \tan 45^{\circ} = \frac{17.5}{d} = 1 \quad\Longrightarrow\quad d = 17.5 \text{ m} \]

At \( 45^{\circ} \) the two legs are always equal, so the distance simply is the height difference. Using either height on its own, or their sum, is where the other options come from.