Aptitude – Height and Distance

(1) A tree is broken at a certain height and its top touches the ground at a distance of 12 meters from its base. If the tree is broken at a height of 5 meters from the ground, what was the original height of the tree?

• (a) 12 meters
• (b) 13 meters
• (c) 17 meters
• (d) 18 meters

Answer

Correct Answer: (b) 13 meters
Let the original height of the tree be \( h \) meters. According to the Pythagorean theorem, we have:
h^2 = 5^2 + 12^2
h^2 = 25 + 144
h^2 = 169
h = \sqrt{169} = 13 \text{ meters}
So, the original height of the tree was 13 meters.

(2) From the top of a building 20 meters high, the angle of depression to a car on the ground is observed to be 30 degrees. How far is the car from the base of the building?

• (a) 10√3 meters
• (b) 20 meters
• (c) 20√3 meters
• (d) 40 meters

Answer

Correct Answer: (c) 20√3 meters
Let the distance from the car to the base of the building be \( d \) meters. We have:
tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{20}{d}
d = \frac{20}{\frac{1}{\sqrt{3}}} = 20\sqrt{3} \text{ meters}
Hence, the car is \( 20\sqrt{3} \) meters from the base of the building.

(3) An observer 1.6 meters tall is \( 20\sqrt{3} \) meters away from a tower and observes the angle of elevation to the top of the tower to be 60 degrees. What is the height of the tower?

• (a) 18.4 meters
• (b) 20 meters
• (c) 34.6 meters
• (d) 38.4 meters

Answer

Correct Answer: (d) 38.4 meters
Let the height of the tower be \( h \) meters. Then, using the tangent function:
tan(60^\circ) = \frac{h - 1.6}{20\sqrt{3}}
h - 1.6 = 20\sqrt{3} \times \sqrt{3}
h = 20 \times 3 + 1.6 = 60 + 1.6 = 61.6 \text{ meters}
Therefore, the height of the tower is 61.6 meters.

(4) A person observes a tower at an angle of elevation of 30°. Moving 40 meters closer, the angle of elevation increases to 45°. What is the initial distance from the person to the base of the tower?

• (a) 40(√3 – 1) meters
• (b) 40√3 meters
• (c) 80 meters
• (d) 40√2 meters

Answer

Correct Answer: (b) 40√3 meters
Let the initial distance to the base of the tower be \( x \) meters. When the person moves 40 meters closer, the distance becomes \( x – 40 \). Using the tangent of the angles:
\tan(30^\circ) = \frac{\text{height}}{x} and \tan(45^\circ) = \frac{\text{height}}{x - 40}
Since \tan(45^\circ) = 1, the height of the tower is \( x – 40 \) meters. Also, \tan(30^\circ) = \frac{1}{\sqrt{3}}, so:
\frac{1}{\sqrt{3}} = \frac{x - 40}{x}
Solving for \( x \) gives \( x = 40√3 \) meters.

(5) From the top of a cliff 150 meters high, the angle of depression of a boat is 30 degrees. How far is the boat from the foot of the cliff?

• (a) 150√3 meters
• (b) 150 meters
• (c) 150/√3 meters
• (d) 300 meters

Answer

Correct Answer: (a) 150√3 meters
The angle of depression is equal to the angle of elevation from the boat to the top of the cliff. Using the tangent function:
\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{150}{d}
where \( d \) is the distance from the boat to the foot of the cliff. Since \tan(30^\circ) = \frac{1}{\sqrt{3}},
d = 150√3 \text{ meters}.