Heights & Distances Questions and Answers

Answer: C

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

\[ d_{30} = h\sqrt{3}, \qquad d_{45} = h \]

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

Rationalising:

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

Numerically about 54.6 m. Multiplying the numerator and denominator by the conjugate is what turns the surd denominator into the plain 2.

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: B

The angle of depression from the tower top equals the angle of elevation of the tower from the bus, so the right triangle has the 80 m tower opposite the 30° angle.

\[ \tan 30^{\circ} = \frac{80}{d} \quad\Longrightarrow\quad d = \frac{80}{\tan 30^{\circ}} = 80\sqrt{3} \]

\[ d = 80 \times 1.732 = 138.56 \approx 138.4 \text{ m} \]

With a 30° depression the ground distance is always \( \sqrt{3} \) times the height - noticeably further than the height itself, so 40 m and 46 m can be discarded on sight.

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: B

The two points are on opposite sides, so the lighthouse foot lies between them and their distances add.

\[ d_{45} = \frac{60}{\tan 45^{\circ}} = 60 \text{ m} \]

\[ d_{60} = \frac{60}{\tan 60^{\circ}} = \frac{60}{\sqrt{3}} = 20\sqrt{3} = 34.64 \text{ m} \]

\[ \text{total} = 60 + 34.64 = 94.64 \approx 94.6 \text{ m} \]

The larger angle gives the shorter distance, as expected - the steeper you have to look, the closer you are.

Answer: A

The shadow is the base of the right triangle and the building is the side opposite the sun's \( 60^{\circ} \) elevation.

\[ \tan 60^{\circ} = \frac{h}{20} \]

\[ h = 20\sqrt{3} = 20 \times 1.732 = 34.64 \text{ m} \]

With the sun above \( 45^{\circ} \) the building must be taller than its shadow, so any answer below 20 m can be discarded immediately.

Answer: D

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

After 10 seconds the depression is \( 30^{\circ} \):

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

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

\[ \text{speed} = \frac{73.2}{10} = 7.32 \text{ m/s} = 7.32 \times \frac{18}{5} = 26.35 \approx 26.28 \text{ km/h} \]

Only the difference of the two distances is travelled in those 10 seconds; using 173.2 m outright would assume the boat set off from the tower.

Answer: C

Read the angle carefully: it is measured from the wall, not from the ground. The distance of the foot from the wall is therefore opposite that 60° angle, so sine applies.

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

\[ x = 10 \sin 60^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \]

\[ x = 5 \times 1.732 = 8.66 \approx 8.65 \text{ m} \]

Assuming the angle is with the ground gives \( 10\cos 60^{\circ} = 5 \) m - which is exactly why 5 m appears among the options.

Answer: A

The pole and its shadow are the two legs of the right triangle formed by the sun's ray.

\[ \tan\theta = \frac{\text{height}}{\text{shadow}} = \frac{18}{6\sqrt{3}} \]

\[ \tan\theta = \frac{3}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3} \]

\[ \theta = 60^{\circ} \]

The shadow being shorter than the pole already tells you the sun is above \( 45^{\circ} \), so 60° is the only standard angle that can fit.

Answer: B

The swimmer's path is the hypotenuse. The width of the river is the perpendicular distance between the parallel banks, which is the side opposite the \( 45^{\circ} \) angle the path makes with the bank.

\[ \sin 45^{\circ} = \frac{\text{width}}{20} \]

\[ \text{width} = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2} = 10 \times 1.414 = 14.14 \text{ m} \]

At \( 45^{\circ} \) the swimmer drifts exactly as far along the bank as the river is wide, so the 20 m swim splits into two equal legs of \( 10\sqrt{2} \) m.