Heights & Distances Questions and Answers

Answer: B

The shadow will be 5 meter long.

The tower's height is the shadow length times the tangent of the Sun's altitude:

\[ \begin{aligned} h &= 15 \tan 30^{\circ} = \frac{15}{\sqrt{3}} = 5\sqrt{3} \\ s &= \frac{h}{\tan 60^{\circ}} = \frac{5\sqrt{3}}{\sqrt{3}} = 5 \end{aligned} \]

Since \(\tan 60^{\circ} = 3 \tan 30^{\circ}\), the shadow at 60° is one-third of the shadow at 30°. 4, 3 and 6 metres do not give that ratio.

Answer: C

The height of the tree is \(16\sqrt{3}\) meter.

The tree, its shadow and the Sun's ray form a right triangle in which the angle of elevation is 60°:

\[ \begin{aligned} \tan 60^{\circ} &= \frac{h}{16} \\ h &= 16\sqrt{3} \approx 27.7 \text{ m} \end{aligned} \]

\(16/\sqrt{3}\) is the answer for an elevation of 30°, where the tree would be shorter than its shadow. 16 m would need an elevation of 45°, and 8 m is half the shadow, which corresponds to no standard angle.

Answer: A

The height of the tower is \(17/\sqrt{3}\) m.

The tower, the ground and the line of sight form a right triangle with the 30° angle at the observer, the 17 m distance as the adjacent side and the height h as the opposite side.

\[ \begin{aligned} \tan 30^{\circ} &= \frac{h}{17} \\ h &= \frac{17}{\sqrt{3}} \approx 9.8\ \text{m} \end{aligned} \]

\(17\sqrt{3}\) would be the height for a 60° angle of elevation, and 17 m would need a 45° angle.

Answer: A

The answer is x = 7.5 metre and y = \(15\sqrt{3}/2\) metre.

The ladder is the hypotenuse, 15 m, making 30° with the ground. The height reached is the side opposite that angle and the distance from the wall the side adjacent to it:

\[ \begin{aligned} x &= 15 \sin 30^\circ = 7.5\ \text{m} \\ y &= 15 \cos 30^\circ = \frac{15\sqrt{3}}{2}\ \text{m} \end{aligned} \]

Option B swaps the two, which would be right only for a 60° angle with the ground. The values with \(\sqrt{2}\) belong to a 45° angle.

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.