Heights & Distances Questions and Answers

Answer: C

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

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

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

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

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: 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: A

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: A

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: D

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

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

\[ \frac{1}{2} = \frac{12.4}{L} \quad\Longrightarrow\quad L = 2 \times 12.4 = 24.8 \text{ m} \]

Since \( \cos 60^{\circ} = \tfrac{1}{2} \), the ladder is always exactly twice the foot distance here. Reaching for \( \tan \) or \( \sin \) instead gives \( 12.4\sqrt{3} \) - the wall height, not the ladder.

Answer: C

Let the tower be \( h \) and let the car's distance be measured from its foot. Because the angle of depression from the top equals the angle of elevation from the car:

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

So in 8 minutes the car covers \( h\sqrt{3} - h = h(\sqrt{3}-1) \), and the distance still to go is \( h \). At constant speed, time is proportional to distance:

\[ t = 8 \times \frac{h}{h(\sqrt{3}-1)} = \frac{8}{\sqrt{3}-1} \]

\[ t = \frac{8(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{8(\sqrt{3}+1)}{2} = 4(\sqrt{3}+1) \]

\[ t = 4 \times 2.732 = 10.93 \text{ min} = 10 \text{ min } 56 \text{ s} \approx 10 \text{ min } 57 \text{ s} \]

The tower height cancels, so it never needed to be known.

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

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

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

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

\[ \tan 30^{\circ} = \frac{15 - h}{5\sqrt{3}} \quad\Longrightarrow\quad 15 - h = \frac{5\sqrt{3}}{\sqrt{3}} = 5 \]

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

Both sightings share the same \( d \); that is what links the two equations.

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.

Answer: B

Both objects are on the same side, so their distances subtract rather than add. Angles of depression from the top equal the angles of elevation from the ground.

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

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

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

The steeper angle belongs to the nearer object. Adding the two distances would be correct only for objects on opposite sides.