Heights & Distances Questions and Answers

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.

Answer: C

The cable has no slack, so it is a straight hypotenuse of 200 m, and the balloon's height is the side opposite the \( 60^{\circ} \) inclination.

\[ \sin 60^{\circ} = \frac{h}{200} \]

\[ h = 200 \times \frac{\sqrt{3}}{2} = 100\sqrt{3} = 173.2 \text{ m} \]

The cable length is a slant distance, so sine applies, not tangent. The height must come out below 200 m, which rules out anything larger on inspection.

Answer: A

The angle of depression from the masthead equals the angle of elevation of the masthead from the boat, so the 210 m height is opposite the \( 30^{\circ} \).

\[ \tan 30^{\circ} = \frac{210}{d} \]

\[ d = \frac{210}{\tan 30^{\circ}} = 210\sqrt{3} \approx 363.7 \text{ m} \]

At \( 30^{\circ} \) the horizontal distance is always \( \sqrt{3} \) times the height, so it must be larger than 210 m - which rules out \( 70\sqrt{3} \) and \( 105\sqrt{3} \) at a glance.

Answer: B

The lighthouse foot lies between the ships, so the total distance is the sum of the two horizontal distances.

\[ d = \frac{h}{\tan 30^{\circ}} + \frac{h}{\tan 45^{\circ}} \]

With \( h = 100 \), \( \tan 30^{\circ} = \tfrac{1}{\sqrt{3}} \) and \( \tan 45^{\circ} = 1 \):

\[ d = 100\sqrt{3} + 100 = 173.2 + 100 = 273.2 \approx 273 \text{ m} \]

Because the ships are on opposite sides the distances add. Had they been on the same side the answer would be the difference, 73 m. Choosing 173 m means only the \( 30^{\circ} \) ship was counted.

Answer: A

The angle of elevation \( \theta \) satisfies

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

With the shadow \( \sqrt{3} \) times the height \( h \):

\[ \tan\theta = \frac{h}{\sqrt{3}\,h} = \frac{1}{\sqrt{3}} \quad\Longrightarrow\quad \theta = 30^{\circ} \]

The pole's height cancels, so no measurement is needed. A shadow longer than the object always means the sun is below \( 45^{\circ} \).

Answer: B

The bent tree splits into a standing stump \( h \) and a bent-over part \( L \), which together made up the original height. The 8 m is the distance from the foot to where the top touches the ground.

\[ \cos 60^{\circ} = \frac{8}{L} \quad\Longrightarrow\quad L = \frac{8}{1/2} = 16 \text{ m} \]

\[ \tan 60^{\circ} = \frac{h}{8} \quad\Longrightarrow\quad h = 8\sqrt{3} \text{ m} \]

\[ \text{original height} = h + L = 8\sqrt{3} + 16 = 8(\sqrt{3} + 2) \text{ m} \]

Numerically that is about 29.9 m. The stump and the bent part must be added - reporting \( 8\sqrt{3} \) alone gives only the part still upright.

Answer: B

F is the foot of the tower, A and B are on the same straight line from it, with \( AF = 200 \) m and \( AB = 80 \) m, so \( BF = 120 \) m.

(a) Height of the tower. From A:

\[ \tan x^{\circ} = \frac{TF}{AF} = \frac{2}{5} \quad\Longrightarrow\quad TF = 200 \times \frac{2}{5} = 80 \text{ m} \]

(b) The angle y. From B, the same 80 m tower is only 120 m away:

\[ \tan y^{\circ} = \frac{80}{120} = \frac{2}{3} = 0.6667 \]

\[ y = \tan^{-1}(0.6667) = 33.69^{\circ} \approx 34^{\circ} \]

Moving 80 m closer raises the angle from about \( 21.8^{\circ} \) to \( 34^{\circ} \) - the elevation grows faster the nearer you get.