Heights & Distances Questions and Answers

Answer: C

Read the angle carefully: it is measured from the wall, not from the ground. The distance of the foot from the wall is therefore opposite that 60° angle, so sine applies.

\[ \sin 60^{\circ} = \frac{\text{distance from wall}}{\text{ladder}} \]

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

\[ x = 5 \times 1.732 = 8.66 \approx 8.65 \text{ m} \]

Assuming the angle is with the ground gives \( 10\cos 60^{\circ} = 5 \) m - which is exactly why 5 m appears among the options.

Answer: B

The angle of depression from the tower top equals the angle of elevation of the tower from the bus, so the right triangle has the 80 m tower opposite the 30° angle.

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

\[ d = 80 \times 1.732 = 138.56 \approx 138.4 \text{ m} \]

With a 30° depression the ground distance is always \( \sqrt{3} \) times the height - noticeably further than the height itself, so 40 m and 46 m can be discarded on sight.

Answer: B

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

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

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

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

The larger angle gives the shorter distance, as expected - the steeper you have to look, the closer you are.

Answer: C

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

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

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

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

Because \( \tan 60^{\circ} / \tan 30^{\circ} = 3 \), the full height is exactly three times the tower - so the flagpole is twice the tower, and 18 m makes the tower 9 m.

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: D

The tree is taller than the wall, which is why the wall's top is seen below the horizontal from the tree top.

From the tree's bottom, the wall top is at 30°:

\[ \tan 30^{\circ} = \frac{11}{d} \quad\Longrightarrow\quad d = 11\sqrt{3} = 19.05 \text{ m} \]

From the tree's top, the wall top is \( H - 11 \) below, at 60°:

\[ \tan 60^{\circ} = \frac{H - 11}{11\sqrt{3}} \quad\Longrightarrow\quad H - 11 = \sqrt{3} \times 11\sqrt{3} = 33 \]

\[ H = 33 + 11 = 44 \text{ m} \]

The tree turns out to be exactly four times the wall's height - a consequence of \( \tan 60^{\circ}\tan 30^{\circ} = 1 \) giving the neat factor of 3.

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: A

Angles are measured from the man's eye, which is 180 cm = 1.8 m above the ground, and he is 5 m from the wall.

Bottom of the window, at 45°:

\[ \text{rise} = 5\tan 45^{\circ} = 5 \text{ m} \quad\Longrightarrow\quad \text{height} = 1.8 + 5 = 6.8 \text{ m} \]

Top of the window, at 60°:

\[ \text{rise} = 5\tan 60^{\circ} = 5\sqrt{3} = 8.66 \text{ m} \quad\Longrightarrow\quad \text{height} = 1.8 + 8.66 = 10.46 \text{ m} \]

\[ \text{window length} = 10.46 - 6.8 = 3.66 \approx 3.65 \text{ m} \]

The 1.8 m appears in both heights and cancels in the subtraction, so the eye height does not actually affect the window's length here - but it is needed if the question asks for either height on its own.

Answer: B

Put the foot of the mountain at the origin. From there the summit is at 45°, so if the mountain is \( H \) km high the summit is at \( (H,\;H) \).

Walking 2 km up a 30° incline moves the observer to

\[ \left(2\cos 30^{\circ},\; 2\sin 30^{\circ}\right) = \left(\sqrt{3},\; 1\right) \]

From this new point the elevation is 60°:

\[ \tan 60^{\circ} = \frac{H - 1}{H - \sqrt{3}} = \sqrt{3} \]

\[ H - 1 = \sqrt{3}H - 3 \quad\Longrightarrow\quad H(1 - \sqrt{3}) = -2 \]

\[ H = \frac{2}{\sqrt{3} - 1} = \frac{2(\sqrt{3}+1)}{2} = \sqrt{3} + 1 = 2.73 \text{ km} \]

The ascent raises the observer as well as bringing them closer, so both coordinates must change - treating the 2 km as purely horizontal is the usual mistake.

Answer: B

Let the tower be \( T \) m high and \( d \) m away. Two sightings:

\[ \tan b = \frac{T}{d} = \frac{4}{5}, \qquad \tan a = \frac{T - 230}{d} = \frac{5}{12} \]

Subtracting eliminates \( T \):

\[ \frac{230}{d} = \frac{4}{5} - \frac{5}{12} = \frac{48 - 25}{60} = \frac{23}{60} \]

\[ d = 230 \times \frac{60}{23} = 600 \text{ m} \]

\[ T = \frac{4}{5} \times 600 = 480 \text{ m}, \qquad T - 230 = 250 \text{ m} \]

The line joining the two tops is the hypotenuse of a right triangle with legs \( d \) and \( T - 230 \):

\[ \sqrt{600^{2} + 250^{2}} = \sqrt{360000 + 62500} = \sqrt{422500} = 650 \text{ m} \]

600 m is the horizontal distance only; the question asks for the slant distance between the tops.