Heights & Distances Questions and Answers

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 leaning tree is the hypotenuse of a right triangle in each case, with the tree's base fixed and its 15 m length unchanged.

Towards A: the top rests 12 m up the wall, so the horizontal gap is

\[ \sqrt{15^{2} - 12^{2}} = \sqrt{225 - 144} = \sqrt{81} = 9 \text{ m} \]

Towards B: the top rests 9 m up, so

\[ \sqrt{15^{2} - 9^{2}} = \sqrt{225 - 81} = \sqrt{144} = 12 \text{ m} \]

The tree stands between the two houses, so the distance between them is the sum:

\[ 9 + 12 = 21 \text{ m} \]

Both are 9-12-15 triangles, the 3-4-5 triple scaled by three.

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: A

Both objects lie on the same side, so their distances subtract. Depression from the top equals elevation from the ground.

\[ d_{45} = \frac{600}{\tan 45^{\circ}} = 600 \text{ m} \]

\[ d_{60} = \frac{600}{\tan 60^{\circ}} = \frac{600}{\sqrt{3}} = 200\sqrt{3} = 346.4 \text{ m} \]

\[ 600 - 346.4 = 253.6 \approx 254 \text{ m} \]

The steeper \( 60^{\circ} \) sighting belongs to the nearer object. Adding the distances would be right only if the objects straddled the tower.

Answer: D

Both aeroplanes lie on the same vertical line, so a single horizontal distance \( d \) serves both sightings.

Upper plane, 900 m high, elevation \( 60^{\circ} \):

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

Lower plane, elevation \( 45^{\circ} \):

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

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

The larger elevation must belong to the higher aircraft, which is what pairs \( 60^{\circ} \) with 900 m.

Answer: B

At \( 45^{\circ} \) the horizontal distance equals the height, so the tower (measured to the man's eye) is \( 100 \) m.

Ten seconds later the depression is \( 30^{\circ} \):

\[ \tan 30^{\circ} = \frac{100}{d} \quad\Longrightarrow\quad d = 100\sqrt{3} = 173.2 \text{ m} \]

\[ \text{distance covered} = 173.2 - 100 = 73.2 \text{ m in } 10 \text{ s} \]

\[ \text{speed} = 7.32 \text{ m/s} = 7.32 \times \frac{18}{5} = 26.35 \text{ km/h} \approx 26.28 \text{ km/h} \]

Use the difference of the two distances, not \( 173.2 \) itself - that would give the speed as if the boat had started at the tower.

Answer: D

Let the tower be \( h \). The two positions give:

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

The 100 m walked is the difference:

\[ h\sqrt{3} - \frac{h}{\sqrt{3}} = 100 \quad\Longrightarrow\quad \frac{3h - h}{\sqrt{3}} = 100 \]

\[ \frac{2h}{\sqrt{3}} = 100 \quad\Longrightarrow\quad h = 50\sqrt{3} \approx 86.6 \text{ m} \]

The pairing \( 30^{\circ}/60^{\circ} \) is convenient because \( \sqrt{3} - \tfrac{1}{\sqrt{3}} \) collapses to \( \tfrac{2}{\sqrt{3}} \).

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: C

The tower is opposite the \( 30^{\circ} \) angle and the distance is adjacent, so:

\[ \tan 30^{\circ} = \frac{100}{d} \quad\Longrightarrow\quad d = \frac{100}{\tan 30^{\circ}} = 100\sqrt{3} \]

\[ d = 100 \times 1.732 = 173.2 \approx 173 \text{ m} \]

A \( 30^{\circ} \) elevation always puts the observer \( \sqrt{3} \) times the height away. Multiplying by \( \tan 30^{\circ} \) instead of dividing gives about 58 m - the classic inversion.

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.