Heights & Distances Questions and Answers

Answer: B

Let the tower be \( T \) m high and \( d \) m away. Two sightings:

\[ \tan b = \frac{T}{d} = \frac{4}{5}, \qquad \tan a = \frac{T - 230}{d} = \frac{5}{12} \]

Subtracting eliminates \( T \):

\[ \frac{230}{d} = \frac{4}{5} - \frac{5}{12} = \frac{48 - 25}{60} = \frac{23}{60} \]

\[ d = 230 \times \frac{60}{23} = 600 \text{ m} \]

\[ T = \frac{4}{5} \times 600 = 480 \text{ m}, \qquad T - 230 = 250 \text{ m} \]

The line joining the two tops is the hypotenuse of a right triangle with legs \( d \) and \( T - 230 \):

\[ \sqrt{600^{2} + 250^{2}} = \sqrt{360000 + 62500} = \sqrt{422500} = 650 \text{ m} \]

600 m is the horizontal distance only; the question asks for the slant distance between the tops.

Answer: B

The balloon rises vertically above \( A \), so its height is the opposite side of a right triangle with base 150 m.

\[ h = 150\tan 60^{\circ} = 150\sqrt{3} = 150 \times 1.732 = 259.8 \text{ m} \]

Two minutes is 120 seconds:

\[ v = \frac{259.8}{120} = 2.165 \approx 2.16 \text{ m/s} \]

Converting the 2 minutes to seconds before dividing is the step most often skipped; leaving it in minutes gives 129.9, which matches no option.

Answer: D

Let \( d \) be the horizontal distance between the building and the tower.

The tower's foot is 30 m below the observer, at a depression of \( 45^{\circ} \):

\[ d = \frac{30}{\tan 45^{\circ}} = 30 \text{ m} \]

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

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

below the building's top, so the tower's height is

\[ 30 - \frac{30}{\sqrt{3}} = 30\left(1 - \frac{1}{\sqrt{3}}\right) \approx 12.7 \text{ m} \]

The tower is shorter than the building, which is consistent with both angles being depressions rather than elevations.

Answer: D

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

\[ \tan\theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3} \]

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

Rationalising \( \tfrac{3}{\sqrt{3}} \) to \( \sqrt{3} \) is the step that makes the standard angle visible; left as a decimal it reads 1.732, which is the same thing.

Answer: B

The angle is measured from the observer's eye, 2 m above the ground, so the trigonometry gives only the part of the tower above eye level.

\[ \tan 30^{\circ} = \frac{x}{10\sqrt{3}} \]

\[ x = 10\sqrt{3} \times \frac{1}{\sqrt{3}} = 10 \text{ m} \]

Adding the observer's own height:

\[ \text{tower} = 10 + 2 = 12 \text{ m} \]

Forgetting to add the 2 m gives 10 m, which is offered as the first distractor. The \( 10\sqrt{3} \) is chosen so the radical cancels exactly.

Answer: B

Put the foot of the mountain at the origin. From there the summit is at 45°, so if the mountain is \( H \) km high the summit is at \( (H,\;H) \).

Walking 2 km up a 30° incline moves the observer to

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

From this new point the elevation is 60°:

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

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

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

The ascent raises the observer as well as bringing them closer, so both coordinates must change - treating the 2 km as purely horizontal is the usual mistake.

Answer: C

Let the poles be \( h \) and \( 2h \). The observer is at the midpoint, so each is 100 m away.

Complementary elevations \( \theta \) and \( 90^{\circ}-\theta \) have tangents whose product is 1, since \( \tan(90^{\circ}-\theta) = \cot\theta \):

\[ \frac{h}{100} \times \frac{2h}{100} = 1 \quad\Longrightarrow\quad 2h^{2} = 10000 \]

\[ h^{2} = 5000 \quad\Longrightarrow\quad h = 50\sqrt{2} = 70.71 \text{ m} \]

\[ 2h = 141.42 \text{ m} \]

So roughly 70.5 m and 141 m. Neither angle is ever needed - the complementarity alone pins the answer down.

Answer: C

The tower is opposite the \( 30^{\circ} \) angle and the distance is adjacent, so:

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

\[ d = 100 \times 1.732 = 173.2 \approx 173 \text{ m} \]

A \( 30^{\circ} \) elevation always puts the observer \( \sqrt{3} \) times the height away. Multiplying by \( \tan 30^{\circ} \) instead of dividing gives about 58 m - the classic inversion.

Answer: A

The building is opposite the \( 30^{\circ} \) altitude and the shadow is the base.

\[ \tan 30^{\circ} = \frac{50}{s} \quad\Longrightarrow\quad s = \frac{50}{\tan 30^{\circ}} \]

\[ s = 50\sqrt{3} \approx 86.6 \text{ m} \]

A low sun casts a long shadow, so the answer must exceed the 50 m height - a quick check that the division went the right way round.

Answer: B

Let AB be the tower and C and D the poles. AB = 70 m. 
In Δ ABC, cot 50º = BC/AB 
Or, cot (90º - 40º) = BC/70 
Or, tan 40º = BC/70 
Or, 0.8391 = BC/70 
Or, BC = 0.8391 × 70 = 58.74 m. 
In Δ ABD, cot 25º = BD/AB 
Or, cot (90º - 65º) = BD/70 
Or, tan 65º = BD/70 
Or, 2.114451 = BD/70 
Or, BD = 2.114451 × 70 = 148.01 m. 
Hence, distance between people = CD = BD - BC 
= 148.01 - 58.74 = 89.27 m.