Heights & Distances Questions and Answers

Answer: B

The angle of elevation \( \theta \) satisfies

\[ \tan\theta = \frac{\text{height of tree}}{\text{length of shadow}} \]

Here the shadow is \( \sqrt{3} \) times the height, so with height \( h \):

\[ \tan\theta = \frac{h}{\sqrt{3}\,h} = \frac{1}{\sqrt{3}} \]

\[ \theta = 30^{\circ} \]

The tree's actual height cancels, so it is never needed. Inverting the ratio gives \( \tan\theta = \sqrt{3} \) and hence 60° - the mirror-image case, where the shadow is shorter than the tree.

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: A

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

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

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

\[ \frac{3h - h}{\sqrt{3}} = 70 \quad\Longrightarrow\quad \frac{2h}{\sqrt{3}} = 70 \]

\[ h = 35\sqrt{3} = 35 \times 1.732 = 60.62 \approx 60.6 \text{ m} \]

70 m is the change in the shadow, not a shadow length - using it directly as one of the two lengths gives 140 m or 35 m, which is why both appear as distractors.

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: 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.