Heights & Distances Questions and Answers

Answer: D

Let the observer be \( d \) from each tower, and let the heights be \( h_{1} \) (at \( 67.5^{\circ} \)) and \( h_{2} \) (at \( 22.5^{\circ} \)).

\[ \frac{h_{1}}{h_{2}} = \frac{d\tan 67.5^{\circ}}{d\tan 22.5^{\circ}} = \frac{\tan 67.5^{\circ}}{\tan 22.5^{\circ}} \]

The two angles are complementary, so

\[ \tan 67.5^{\circ} = \tan(90^{\circ} - 22.5^{\circ}) = \cot 22.5^{\circ} = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1 \]

\[ \frac{h_{1}}{h_{2}} = \frac{\sqrt{2}+1}{\sqrt{2}-1} = (\sqrt{2}+1)^{2} = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2} \]

So the ratio is \( (3 + 2\sqrt{2}) : 1 \), about \( 5.83 : 1 \). The distance \( d \) cancels, which is why "exactly midway" is all the question needs to say.

Answer: C

Correct answer: \(30\sqrt{3}\) m

The height and the base distance are the opposite and adjacent sides of the angle of elevation.

\[\tan 60^\circ = \frac{h}{30} \quad\Longrightarrow\quad h = 30\tan 60^\circ = 30\sqrt{3} \approx 51.96 \text{ m}\]

Answer: A

Both aircraft are on the same vertical line, so they share one horizontal distance \( d \) from the observer.

For the upper plane at 900 m, elevation 60°:

\[ \tan 60^{\circ} = \frac{900}{d} \quad\Longrightarrow\quad d = \frac{900}{\sqrt{3}} = 300\sqrt{3} = 519.6 \text{ m} \]

For the lower plane, elevation 45°:

\[ h = d\tan 45^{\circ} = 519.6 \text{ m} \]

\[ \text{difference} = 900 - 519.6 = 380.4 \approx 381 \text{ m} \]

The steeper 60° sighting must belong to the higher aircraft, which fixes which angle goes with which plane.

Answer: D

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: A

The balloon rises straight up from \( A \), so its height after 2 minutes is the opposite side of a right triangle with base 150 m.

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

Two minutes is 120 seconds, so

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

Converting the time to seconds before dividing is the step most often missed; dividing by 2 instead gives 129.9 m per minute, which matches none of the options.

Answer: C

Guddi's eye, the top of the tree and the top of the mall lie on one straight line, so the tree and the mall are the vertical sides of two similar right triangles sharing the apex at Guddi.

\[ \frac{\text{tree height}}{\text{distance to tree}} = \frac{\text{mall height}}{\text{distance to mall}} \]

\[ \frac{10}{20} = \frac{H}{1000} \]

\[ H = 1000 \times \frac{10}{20} = 1000 \times 0.5 = 500 \text{ m} \]

The 1000 m is measured from Guddi, not from the tree, so it is the full base of the larger triangle - using 980 m here is the usual slip.

Answer: C

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

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

\[ d = \frac{200}{1/\sqrt{3}} = 200\sqrt{3} \]

\[ d = 200 \times 1.732 = 346.4 \text{ m} \]

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: C

The angle of depression from the tower top equals the angle of elevation of the top from the bus, so the 80 m height lies opposite the \( 30^{\circ} \).

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

\[ d = 80 \times 1.73 = 138.4 \text{ m} \]

At \( 30^{\circ} \) the ground distance is always \( \sqrt{3} \) times the height, so it must exceed 80 m - a useful sanity check before computing.

Answer: D

Join the two chimney tops. The horizontal leg of that line is the gap between the houses, and its vertical leg is the difference in their heights:

\[ 25 - 10 = 15 \text{ m} \]

The line makes 45° with the ground, and \( \tan 45^{\circ} = 1 \), so the two legs are equal:

\[ \tan 45^{\circ} = \frac{15}{d} = 1 \quad\Longrightarrow\quad d = 15 \text{ m} \]

The trap is using the sum 35 m or one height alone; only the height difference lies opposite the 45° angle.

Answer: C

The sun's rays hit the man and the pole at the same angle, so the two object–shadow triangles are similar and the ratio of height to shadow is the same for both.

\[ \frac{\text{man's height}}{\text{man's shadow}} = \frac{\text{pole's height}}{\text{pole's shadow}} \]

\[ \frac{6}{4} = \frac{h}{36} \]

\[ h = 36 \times \frac{6}{4} = 36 \times 1.5 = 54 \text{ ft} \]

The same result by scale factor: the pole's shadow is \( 36 \div 4 = 9 \) times the man's, so the pole is 9 times his height, \( 9 \times 6 = 54 \) ft. Multiplying 6 by 6 instead treats the shadow ratio as \( 36 \div 6 \), which mixes a height with a length.