Heights & Distances Questions and Answers

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.

Answer: D

Let the pole be \( h \) m high and let \( d \) be the horizontal gap between the tower and the pole.

From the pole's foot, the tower top is at \( 60^{\circ} \):

\[ \tan 60^{\circ} = \frac{15}{d} \quad\Longrightarrow\quad d = \frac{15}{\sqrt{3}} = 5\sqrt{3} \approx 8.66 \text{ m} \]

From the pole's top, the tower top is only \( 15 - h \) above, at \( 30^{\circ} \):

\[ \tan 30^{\circ} = \frac{15 - h}{5\sqrt{3}} \quad\Longrightarrow\quad 15 - h = \frac{5\sqrt{3}}{\sqrt{3}} = 5 \]

\[ h = 10 \text{ m} \]

Both sightings share the same \( d \); that is what links the two equations.

Answer: C

Both ships lie on the same side of the helicopter, so the distance between them is the difference of their horizontal distances from the point directly below it.

\[ d_{60} = \frac{1200}{\tan 60^{\circ}} = \frac{1200}{\sqrt{3}} = 400\sqrt{3} = 692.8 \text{ m} \]

\[ d_{45} = \frac{1200}{\tan 45^{\circ}} = 1200 \text{ m} \]

\[ \text{separation} = 1200 - 400\sqrt{3} = 400\left(3 - \sqrt{3}\right) \]

\[ = 400 \times 1.268 = 507.2 \text{ m} \]

The steeper \( 60^{\circ} \) depression belongs to the nearer ship. Stopping at \( 400\sqrt{3} \) gives that ship's distance from the helicopter's base, not the gap between the two.

Answer: D

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

Complementary angles mean the two elevations are \( \theta \) and \( 90^{\circ} - \theta \), and since \( \tan(90^{\circ}-\theta) = \cot\theta \), their tangents multiply to 1:

\[ \frac{h}{100} \times \frac{2h}{100} = 1 \]

\[ 2h^{2} = 10000 \quad\Longrightarrow\quad h^{2} = 5000 \quad\Longrightarrow\quad h = 50\sqrt{2} = 70.71 \text{ m} \]

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

So the poles are about 70.5 m and 141 m. The identity \( \tan\theta\cot\theta = 1 \) is what makes the problem solvable without knowing either angle.

Answer: C

Correct answer: 13 m

The line joining the tops is the hypotenuse of a right triangle whose horizontal side is the gap and whose vertical side is the difference in heights.

\[d = \sqrt{12^2 + (11-6)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ m}\]

It is the difference of the heights that matters, not their sum.