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Heights & Distances Questions and Answers

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61.

Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30 degree and 45 degree respectively. If the lighthouse is 100 m high, the distance between the two ships is

Answer: A

The ships lie on opposite sides, so the lighthouse foot is between them and the two horizontal distances add.

\[ d_{30} = \frac{100}{\tan 30^{\circ}} = 100\sqrt{3} = 173.2 \text{ m} \]

\[ d_{45} = \frac{100}{\tan 45^{\circ}} = 100 \text{ m} \]

\[ \text{distance apart} = 173.2 + 100 = 273.2 \approx 273 \text{ m} \]

Had the ships been on the same side the answer would be the difference, 73 m. Answering 173 m counts only the \( 30^{\circ} \) ship.

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62.

The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°,the elevation changes to 60°. What is the approximate height of the mou

Answer: C

Put the mountain's foot at the origin. The summit is seen at \( 45^{\circ} \) from there, so for a height \( H \) km the summit sits at \( (H,\;H) \).

Climbing 2 km up a \( 30^{\circ} \) incline moves the observer to

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

From there the elevation is \( 60^{\circ} \):

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

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

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

The ascent lifts the observer as well as moving them closer, so both coordinates change - treating the 2 km as horizontal only is the standard error.

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63.

From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is \( 30^{\circ} \). The angle of elevation of the top of a water tank standing on the tower is \( 45^{\circ} \). Find the height of the tower and the height of the water tank.

Answer: B

Both sightings are taken from the same point 40 m away, so they share one base.

Tower, at \( 30^{\circ} \):

\[ h = 40\tan 30^{\circ} = \frac{40}{\sqrt{3}} = 23.09 \approx 23.1 \text{ m} \]

Top of the tank, at \( 45^{\circ} \):

\[ H = 40\tan 45^{\circ} = 40 \text{ m} \]

The tank stands on the tower, so its own height is the difference:

\[ 40 - 23.09 = 16.91 \approx 16.9 \text{ m} \]

The second angle measures to the top of the tank, giving the combined height - reporting 40 m as the tank's height is the error the distractors target.

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64.

An aeroplane flying at a height of 4000 m passes vertically above another aeroplane at the instant when the angles of elevation of the two planes from the same point on the ground are \( 60^{\circ} \) and \( 45^{\circ} \) respectively. Find the vertical distance between the two planes. (Take \( \sqrt{3} = 1.732 \))

Answer: C

Both planes lie on the same vertical line, so one horizontal distance \( d \) serves both sightings. The higher plane takes the steeper angle.

\[ \tan 60^{\circ} = \frac{4000}{d} \quad\Longrightarrow\quad d = \frac{4000}{1.732} = 2309.47 \text{ m} \]

\[ h = d\tan 45^{\circ} = 2309.47 \text{ m} \]

\[ \text{vertical gap} = 4000 - 2309.47 = 1690.53 \text{ m} \]

At \( 45^{\circ} \) the lower plane's height simply equals the horizontal distance, which is why only one division is really needed.

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65.

A ladder leaning against a vertical wall makes an angle of \( 45^{\circ} \) with the level ground. The foot of the ladder is 3 m from the wall. Find the length of the ladder.

Answer: A

The ladder is the hypotenuse and the 3 m is adjacent to the \( 45^{\circ} \) angle, so cosine relates them.

\[ \cos 45^{\circ} = \frac{3}{L} \quad\Longrightarrow\quad \frac{1}{\sqrt{2}} = \frac{3}{L} \]

\[ L = 3\sqrt{2} = 3 \times 1.414 = 4.243 \approx 4.23 \text{ m} \]

At \( 45^{\circ} \) the ladder reaches 3 m up the wall as well, so it is the diagonal of a 3 m square - hence the factor \( \sqrt{2} \).

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66.

When the sun's altitude changes from 30° to 60°, the length of the shadow of a tower decreases by 70m. What is the height of the tower?

Answer: B

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

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

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

\[ \frac{2h}{\sqrt{3}} = 70 \quad\Longrightarrow\quad h = 35\sqrt{3} = 60.62 \approx 60.6 \text{ m} \]

70 m is the decrease, not a shadow length. Mistaking it for one gives either 140 m or 35 m - hence those distractors.

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67.

On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 600 m, the distance between the objects is approximately equal to :

Answer: B

Both objects are on the same side, so their distances subtract rather than add. Angles of depression from the top equal the angles of elevation from the ground.

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

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

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

The steeper angle belongs to the nearer object. Adding the two distances would be correct only for objects on opposite sides.

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68.

If the angle of elevation of cloud from a point 200 m above a lake is 30º and the angle of depression of its reflection in the lake is 60º, then find the height of the cloud above the lake.

Answer: A

Let the cloud be \( H \) m above the lake surface, and let \( d \) be the horizontal distance from the observation point. The observer is 200 m above the lake, and the reflection sits \( H \) m below the surface.

Relative to the observer, the cloud is \( H - 200 \) up and the reflection is \( H + 200 \) down:

\[ \tan 30^{\circ} = \frac{H - 200}{d}, \qquad \tan 60^{\circ} = \frac{H + 200}{d} \]

Dividing eliminates \( d \):

\[ \frac{\tan 60^{\circ}}{\tan 30^{\circ}} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 = \frac{H+200}{H-200} \]

\[ H + 200 = 3H - 600 \quad\Longrightarrow\quad 2H = 800 \quad\Longrightarrow\quad H = 400 \text{ m} \]

The reflection is the mirror image in the water surface, so its depth below the surface equals the cloud's height above it - that is why the second numerator is \( H + 200 \) and not \( H - 200 \).

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69.

Two persons are on either sides of a tower of height 50 m. The persons observers the top of the tower at an angle of elevation of 30° and 60°. If a car crosses these two persons in 10 seconds, what is

Answer: D

The two observers are on opposite sides, so the car's path is the sum of the two horizontal distances from the 50 m tower.

\[ d = \frac{50}{\tan 30^{\circ}} + \frac{50}{\tan 60^{\circ}} = 50\sqrt{3} + \frac{50}{\sqrt{3}} \]

\[ d = \frac{150 + 50}{\sqrt{3}} = \frac{200}{\sqrt{3}} = \frac{200\sqrt{3}}{3} \text{ m} \]

That is covered in 10 seconds, so

\[ v = \frac{200\sqrt{3}}{3 \times 10} = \frac{20\sqrt{3}}{3} \text{ m/s} \]

\[ v = \frac{20\sqrt{3}}{3} \times \frac{18}{5} = \frac{360\sqrt{3}}{15} = 24\sqrt{3} \text{ km/h} \]

Numerically that is about 41.6 km/h. Converting m/s to km/h with the factor \( \tfrac{18}{5} \) is the step the distractors are built around.

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70.

A man observes the elevation of a balloon to be \( 45^{\circ} \) at a point A. He then walks towards the balloon and at a point B finds the elevation to be \( 60^{\circ} \). Walking further in the same direction he finds the balloon directly overhead at a height of 450 m. The distance travelled from A to B is

Answer: A

The balloon ends up directly overhead at 450 m, so measure both earlier observations from that vertical line.

From A, at \( 45^{\circ} \):

\[ d_{A} = \frac{450}{\tan 45^{\circ}} = 450 \text{ m} \]

From B, at \( 60^{\circ} \):

\[ d_{B} = \frac{450}{\tan 60^{\circ}} = \frac{450}{\sqrt{3}} = 150\sqrt{3} = 259.8 \text{ m} \]

\[ AB = d_{A} - d_{B} = 450 - 150\sqrt{3} = 150\left(3 - \sqrt{3}\right) \]

\[ = 150 \times 1.268 = 190.2 \text{ m} \]

The balloon is assumed to hold its height while the man walks, so all three sightings share the same 450 m - that is what makes the two distances comparable.

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