Heights & Distances Questions and Answers

Answer: A

Let the lower part be \( h \), so the upper part is \( 9h \) and the pole is \( 10h \). From a point 15 m away, let \( \alpha \) be the angle subtended by the lower part. The upper part subtends the same angle, so the whole pole subtends \( 2\alpha \).

\[ \tan\alpha = \frac{h}{15}, \qquad \tan 2\alpha = \frac{10h}{15} \]

Applying the double-angle identity \( \tan 2\alpha = \dfrac{2\tan\alpha}{1-\tan^{2}\alpha} \) with \( t = \dfrac{h}{15} \):

\[ 10t = \frac{2t}{1 - t^{2}} \quad\Longrightarrow\quad 10(1 - t^{2}) = 2 \]

\[ t^{2} = \frac{8}{10} = \frac{4}{5} \quad\Longrightarrow\quad \frac{h^{2}}{225} = \frac{4}{5} \quad\Longrightarrow\quad h^{2} = 180 \]

\[ h = 6\sqrt{5} \quad\Longrightarrow\quad \text{pole} = 10h = 60\sqrt{5} \text{ m} \]

The key step is recognising that "equal angles" makes the total angle exactly double the lower one, which is what lets the double-angle formula in.

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

Let the tower be \( h \) m. The shadow shortens as the sun rises.

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

\[ h\sqrt{3} - \frac{h}{\sqrt{3}} = 70 \quad\Longrightarrow\quad \frac{3h - h}{\sqrt{3}} = 70 \]

\[ \frac{2h}{\sqrt{3}} = 70 \quad\Longrightarrow\quad h = 35\sqrt{3} = 60.62 \approx 60.6 \text{ m} \]

70 m is the decrease, not a shadow length. Mistaking it for one gives either 140 m or 35 m - hence those distractors.

Answer: C

The ladder is the hypotenuse; the 12.4 m ground gap is adjacent to the \( 60^{\circ} \) elevation, so cosine relates them.

\[ \cos 60^{\circ} = \frac{12.4}{L} \quad\Longrightarrow\quad \frac{1}{2} = \frac{12.4}{L} \]

\[ L = 2 \times 12.4 = 24.8 \text{ m} \]

Because \( \cos 60^{\circ} = \tfrac{1}{2} \), the ladder is always exactly twice the foot distance at this angle. Using \( \tan \) instead returns \( 12.4\sqrt{3} = 21.5 \) m, which is the height reached up the wall.

Answer: C

Angles of depression from the top equal the angles of elevation from the car, so for a tower of height \( h \):

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

In 8 minutes the car covers \( h\sqrt{3} - h \); the distance left is \( h \). At constant speed time is proportional to distance:

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

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

The height \( h \) cancels, so the tower's size is irrelevant - which is why the question can omit it.

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: 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: 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: C

Put the mountain's foot at the origin. The summit is seen at \( 45^{\circ} \) from there, so for a height \( H \) km the summit sits at \( (H,\;H) \).

Climbing 2 km up a \( 30^{\circ} \) incline moves the observer to

\[ \left(2\cos 30^{\circ},\; 2\sin 30^{\circ}\right) = \left(\sqrt{3},\; 1\right) \]

From there the elevation is \( 60^{\circ} \):

\[ \frac{H - 1}{H - \sqrt{3}} = \tan 60^{\circ} = \sqrt{3} \]

\[ H - 1 = \sqrt{3}H - 3 \quad\Longrightarrow\quad H(\sqrt{3} - 1) = 2 \]

\[ H = \frac{2}{\sqrt{3}-1} = \sqrt{3} + 1 = 2.73 \text{ km} \]

The ascent lifts the observer as well as moving them closer, so both coordinates change - treating the 2 km as horizontal only is the standard error.

Answer: D

The two observers are on opposite sides, so the car's path is the sum of the two horizontal distances from the 50 m tower.

\[ d = \frac{50}{\tan 30^{\circ}} + \frac{50}{\tan 60^{\circ}} = 50\sqrt{3} + \frac{50}{\sqrt{3}} \]

\[ d = \frac{150 + 50}{\sqrt{3}} = \frac{200}{\sqrt{3}} = \frac{200\sqrt{3}}{3} \text{ m} \]

That is covered in 10 seconds, so

\[ v = \frac{200\sqrt{3}}{3 \times 10} = \frac{20\sqrt{3}}{3} \text{ m/s} \]

\[ v = \frac{20\sqrt{3}}{3} \times \frac{18}{5} = \frac{360\sqrt{3}}{15} = 24\sqrt{3} \text{ km/h} \]

Numerically that is about 41.6 km/h. Converting m/s to km/h with the factor \( \tfrac{18}{5} \) is the step the distractors are built around.