Heights & Distances Questions and Answers

Answer: C

Let \( d \) be the horizontal distance from the hill to the pole.

The pole's bottom is 100 m below the observer, at a depression of \( 60^{\circ} \):

\[ d = \frac{100}{\tan 60^{\circ}} = \frac{100}{\sqrt{3}} = 57.74 \text{ m} \]

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

\[ d\tan 30^{\circ} = \frac{100}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{100}{3} = 33.33 \text{ m} \]

below the hilltop, so the pole's height is

\[ 100 - 33.33 = 66.67 \text{ m} \]

The smaller depression belongs to the higher point, being nearer the observer's eye level.

Answer: B

"Increases by \( 15^{\circ} \)" means the new elevation is \( 30^{\circ} + 15^{\circ} = 45^{\circ} \), not \( 15^{\circ} \).

\[ d_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}, \qquad d_{45} = \frac{h}{\tan 45^{\circ}} = h \]

The 40 m walked is the difference:

\[ h\sqrt{3} - h = 40 \quad\Longrightarrow\quad h = \frac{40}{\sqrt{3}-1} = \frac{40(\sqrt{3}+1)}{2} = 20(\sqrt{3}+1) \]

\[ h = 20 \times 2.732 = 54.64 \approx 54.6 \text{ m} \]

Rationalising the denominator by multiplying top and bottom by \( \sqrt{3}+1 \) keeps the arithmetic exact until the final step.

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

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

The ladder is the hypotenuse and the 4.6 m ground gap is adjacent to the \( 60^{\circ} \) elevation, so cosine is the ratio.

\[ \cos 60^{\circ} = \frac{4.6}{L} \quad\Longrightarrow\quad \frac{1}{2} = \frac{4.6}{L} \]

\[ L = 2 \times 4.6 = 9.2 \text{ m} \]

Since \( \cos 60^{\circ} = \tfrac{1}{2} \), the ladder is exactly twice the foot distance. Using tangent instead returns \( 4.6\sqrt{3} = 7.97 \) m, which is the height reached on the wall - hence the 7.8 m distractor.

Answer: B

The shadow lengthens as the sun drops from \( 45^{\circ} \) to \( 30^{\circ} \).

\[ s_{45} = \frac{h}{\tan 45^{\circ}} = h, \qquad s_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3} \]

\[ h\sqrt{3} - h = 10 \quad\Longrightarrow\quad h(\sqrt{3}-1) = 10 \]

\[ h = \frac{10}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{10(\sqrt{3}+1)}{2} = 5(\sqrt{3}+1) \]

\[ h = 5 \times 2.732 = 13.66 \text{ m} \]

The 10 m is the increase in shadow, not a shadow length in its own right.

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: A

Let the pole be \( h \) m and \( d \) the horizontal gap between tower and pole.

From the pole's foot, the 15 m tower top is at \( 60^{\circ} \):

\[ d = \frac{15}{\tan 60^{\circ}} = \frac{15}{\sqrt{3}} = 5\sqrt{3} \approx 8.66 \text{ m} \]

From the pole's top the tower top is only \( 15 - h \) higher, at \( 30^{\circ} \):

\[ 15 - h = d\tan 30^{\circ} = 5\sqrt{3} \times \frac{1}{\sqrt{3}} = 5 \]

\[ h = 10 \text{ m} \]

The shared distance \( d \) is what ties the two sightings together.

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: D

The elevation rises from \( 30^{\circ} \) to \( 30^{\circ} + 15^{\circ} = 45^{\circ} \). Writing the two horizontal distances in terms of the height \( h \):

\[ d_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}, \qquad d_{45} = \frac{h}{\tan 45^{\circ}} = h \]

The observer walked the difference, 40 m:

\[ h\sqrt{3} - h = 40 \quad\Longrightarrow\quad h(\sqrt{3}-1) = 40 \]

\[ h = \frac{40}{\sqrt{3}-1} = \frac{40(\sqrt{3}+1)}{2} = 20(\sqrt{3}+1) \]

\[ h = 20 \times 2.732 = 54.64 \approx 54.6 \text{ m} \]

"Increases by 15°" means the new angle is 45°, not 15° - reading it as the latter is the trap.