Quantitative Aptitude — Heights and Distances

Let the height of the tower be 'h' and the distance from the foot of the tower to the point on the ground be 'd'. Given: d = 60 meters, Angle of elevation = 30 degrees. We use the tangent function: tan(angle) = opposite/adjacent tan(30 degrees) = h / d 1/sqrt(3) = h / 60 h = 60 / sqrt(3) h = 60*sqrt(3) / 3 h = 20*sqrt(3) meters.

Let the height of the lighthouse be 'h' and the distance of the ship from the lighthouse be 'd'. Given: h = 75 meters, Angle of depression = 45 degrees. The angle of depression is equal to the angle of elevation from the ship to the top of the lighthouse. So, the angle of elevation = 45 degrees. We use the tangent function: tan(angle) = opposite/adjacent tan(45 degrees) = h / d 1 = 75 / d d = 75 meters.

Let the height of the pole be 'h' and the length of its shadow be 's'. Let the angle of elevation of the sun be 'theta'. Given: h = 10 meters, s = 10*sqrt(3) meters. We use the tangent function: tan(theta) = opposite/adjacent tan(theta) = h / s tan(theta) = 10 / (10*sqrt(3)) tan(theta) = 1 / sqrt(3) Therefore, theta = 30 degrees.

Let the height of the tower be 'h'. Let the two points on the ground be C and D, and the base of the tower be B. Let D be closer to the tower. Angle of elevation from D = 60 degrees. Angle of elevation from C = 30 degrees. Distance CD = 40 meters. In right triangle ADB (where A is the top of the tower): tan(60) = h / BD => sqrt(3) = h / BD => BD = h / sqrt(3) In right triangle ACB: tan(30) = h / BC => 1/sqrt(3) = h / (BD + CD) 1/sqrt(3) = h / (h/sqrt(3) + 40) h/sqrt(3) + 40 = h*sqrt(3) 40 = h*sqrt(3) - h/sqrt(3) 40 = h * (3 - 1) / sqrt(3) 40 = 2h / sqrt(3) h = (40 * sqrt(3)) / 2 h = 20*sqrt(3) meters.

Let the height of the chimney be H. The observer's height is 1.5 m. The distance from the observer to the chimney is 28.5 m. The angle of elevation from her eyes to the top of the chimney is 45 degrees. Let the height of the chimney above the observer's eye level be h'. In the right triangle formed: tan(45 degrees) = h' / 28.5 1 = h' / 28.5 h' = 28.5 meters. The total height of the chimney H = h' + observer's height H = 28.5 + 1.5 H = 30 meters.

Let the height of the building be H_b = 60 m. Let the height of the tower be H_t. The horizontal distance between them is d = 60*sqrt(3) m. The angle of elevation from the top of the building to the top of the tower is 30 degrees. Consider a right triangle formed by the top of the building, the top of the tower, and a point directly above the base of the tower at the height of the building. Let the height of the tower above the building be h'. tan(30 degrees) = h' / d 1/sqrt(3) = h' / (60*sqrt(3)) h' = (60*sqrt(3)) / sqrt(3) h' = 60 meters. The total height of the tower H_t = H_b + h' H_t = 60 + 60 H_t = 120 meters.

Height of the building = 31.6 m. Height of the boy = 1.6 m. Effective height of the building from the boy's eye level = 31.6 - 1.6 = 30 m. Let the initial distance from the building be x1 and the final distance be x2. Initial angle of elevation = 30 degrees. Final angle of elevation = 60 degrees. In the first case: tan(30) = 30 / x1 => 1/sqrt(3) = 30 / x1 => x1 = 30*sqrt(3) meters. In the second case: tan(60) = 30 / x2 => sqrt(3) = 30 / x2 => x2 = 30 / sqrt(3) = 10*sqrt(3) meters. The distance walked by the boy = x1 - x2 Distance walked = 30*sqrt(3) - 10*sqrt(3) Distance walked = 20*sqrt(3) meters.

Let the length of the ladder be 'L' and the height it reaches on the wall be 'h'. Let the angle made by the ladder with the ground be 'theta'. Given: L = 10 meters, h = 5*sqrt(3) meters. We use the sine function: sin(theta) = opposite/hypotenuse sin(theta) = h / L sin(theta) = (5*sqrt(3)) / 10 sin(theta) = sqrt(3) / 2 Therefore, theta = 60 degrees.

Let the height of the tower be 'h'. Let the two points on the ground be C and D, on opposite sides of the tower's base B. Angle of elevation from C = 30 degrees. Angle of elevation from D = 45 degrees. Distance CD = 100 meters. In right triangle ABC: tan(30) = h / BC => 1/sqrt(3) = h / BC => BC = h*sqrt(3) In right triangle ABD: tan(45) = h / BD => 1 = h / BD => BD = h We know that CD = BC + BD. 100 = h*sqrt(3) + h 100 = h(sqrt(3) + 1) h = 100 / (sqrt(3) + 1) To rationalize the denominator, multiply numerator and denominator by (sqrt(3) - 1): h = 100 * (sqrt(3) - 1) / ((sqrt(3) + 1) * (sqrt(3) - 1)) h = 100 * (sqrt(3) - 1) / (3 - 1) h = 100 * (sqrt(3) - 1) / 2 h = 50(sqrt(3) - 1) meters.

Let the height of the cliff be 'h'. Let the initial position of the boat be C and the position after 8 minutes be D. Let the base of the cliff be B. Angle of depression from C = 30 degrees. (So, angle of elevation from C to top of cliff = 30 degrees) Angle of depression from D = 60 degrees. (So, angle of elevation from D to top of cliff = 60 degrees) In right triangle ABC: tan(30) = h / BC => 1/sqrt(3) = h / BC => BC = h*sqrt(3) In right triangle ABD: tan(60) = h / BD => sqrt(3) = h / BD => BD = h / sqrt(3) The distance covered by the boat in 8 minutes is CD = BC - BD. CD = h*sqrt(3) - h/sqrt(3) = h * (3 - 1) / sqrt(3) = 2h / sqrt(3). Speed of the boat = Distance / Time = (2h/sqrt(3)) / 8 = h / (4*sqrt(3)) meters per minute. Time taken to travel from D to B (shore) = Distance BD / Speed Time = (h/sqrt(3)) / (h / (4*sqrt(3))) Time = (h/sqrt(3)) * (4*sqrt(3) / h) Time = 4 minutes. Total time taken by the boat to reach the shore from the initial point C = Time (C to D) + Time (D to B) Total time = 8 minutes + 4 minutes = 12 minutes.