Heights & Distances Questions and Answers

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 elevation is given as an inverse sine, so convert it to a tangent before using it with the horizontal distance.

\[ \sin\theta = \frac{3}{5} \quad\Longrightarrow\quad \cos\theta = \sqrt{1 - \left(\tfrac{3}{5}\right)^{2}} = \frac{4}{5} \]

\[ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3}{4} \]

\[ h = 20\tan\theta = 20 \times \frac{3}{4} = 15 \text{ m} \]

This is the 3–4–5 triangle. Using \( \tfrac{3}{5} \) directly with the 20 m base gives 12 m, the error the question is testing for.

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: A

Both objects lie on the same side, so their distances subtract. Depression from the top equals elevation from the ground.

\[ d_{45} = \frac{600}{\tan 45^{\circ}} = 600 \text{ m} \]

\[ d_{60} = \frac{600}{\tan 60^{\circ}} = \frac{600}{\sqrt{3}} = 200\sqrt{3} = 346.4 \text{ m} \]

\[ 600 - 346.4 = 253.6 \approx 254 \text{ m} \]

The steeper \( 60^{\circ} \) sighting belongs to the nearer object. Adding the distances would be right only if the objects straddled the tower.

Answer: A

The \( 60^{\circ} \) is measured from the wall, so the ground distance is opposite that angle and sine is the right ratio.

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

\[ x = 5 \times 1.73 = 8.65 \text{ m} \]

Reading the angle as being with the ground gives \( 10\cos 60^{\circ} = 5 \) m instead - a different problem entirely.

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: B

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

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

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

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

The larger angle yields the shorter distance: the steeper you must look, the closer you stand.

Answer: C

Let \( d \) be the horizontal distance from the hill to the pole.

The pole's bottom is 100 m below the observer, at a depression of \( 60^{\circ} \):

\[ d = \frac{100}{\tan 60^{\circ}} = \frac{100}{\sqrt{3}} = 57.74 \text{ m} \]

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

\[ d\tan 30^{\circ} = \frac{100}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{100}{3} = 33.33 \text{ m} \]

below the hilltop, so the pole's height is

\[ 100 - 33.33 = 66.67 \text{ m} \]

The smaller depression belongs to the higher point, being nearer the observer's eye level.

Answer: A

Let the tower be \( h \) m and the observer \( d \) m away. The flagpole's bottom is the tower's top.

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

\[ \tan 60^{\circ} = \frac{h+18}{d} \quad\Longrightarrow\quad h + 18 = \sqrt{3} \cdot h\sqrt{3} = 3h \]

\[ 2h = 18 \quad\Longrightarrow\quad h = 9 \text{ m} \]

Since \( \tan 60^{\circ} \div \tan 30^{\circ} = 3 \), the tower-plus-flagpole is always three times the tower alone - so the flagpole equals twice the tower.

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.