Quantitative Aptitude — Trigonometry

1 + tan²θ = 1 + sin²θ/cos²θ = (cos²θ + sin²θ)/cos²θ = 1/cos²θ = sec²θ.

Height difference = 17 - 12 = 5m. Horizontal distance = 12m. Using Pythagoras theorem: √(12² + 5²) = √(144 + 25) = √169 = 13m.

Let the distance from the wall be 'Base'. cos 60° = Base/Hypotenuse. So, Base = 10 × cos 60° = 10 × 1/2 = 5 m.

(sin θ + cos θ)² = sin²θ + cos²θ + 2 sin θ cos θ. Since sin²θ + cos²θ = 1, the expression becomes 1 + 2 sin θ cos θ.

If sin θ = 1/√2, then θ = 45°. So, tan 45° + cos 45° = 1 + 1/√2 = (√2 + 1)/√2.

This is a basic trigonometric identity: 1 + cot²θ = cosec²θ.

Let θ be the angle of elevation. tan θ = Height/Base = 10 / (10√3) = 1/√3. Therefore, θ = 30°.

We know that sin 30° = 1/2 and cos 60° = 1/2. So, sin 30° + cos 60° = 1/2 + 1/2 = 1.

If tan θ = Perpendicular/Base = 3/4, then using Pythagoras theorem, Hypotenuse = √(3² + 4²) = √(9 + 16) = √25 = 5. So, sin θ = Perpendicular/Hypotenuse = 3/5.

We know that sin 30° = 1/2 and cos 60° = 1/2. So, sin 30° / cos 60° = (1/2) / (1/2) = 1.

From 5cos A + 3 = 0, we get cos A = -3/5. Since A lies in the third quadrant, sin A is negative. Using sin²A + cos²A = 1, sin A = -√(1 - (-3/5)²) = -√(1 - 9/25) = -√(16/25) = -4/5. We know from trigonometric identities that cot²A = cosec²A - 1, which simplifies the last part of the expression: cot²A - (cosec²A - 1) = 0. Substituting the values into the remaining expression gives 10(-4/5) + 25(-3/5) = -8 - 15 = -23.

The length of a tangent from a point (x1, y1) to a circle is given by √S11. Substituting the point (1, 2) into the circle equation gives √(1² + 2² - 2(1) + 3(2) + k) = √10. Squaring both sides, we get 1 + 4 - 2 + 6 + k = 10, which simplifies to 9 + k = 10. Therefore, k = 10 - 9 = 1.

First, find the slope (m1) of the line joining (2, 3) and (5, 6), which is (6 - 3) / (5 - 2) = 3 / 3 = 1. The slope (m) of a line perpendicular to it satisfies m × m1 = -1, so m = -1. Using the point-slope form for a line passing through (4, 1) with slope -1, the equation is y - 1 = -1(x - 4), which simplifies to y - 1 = -x + 4, or x + y - 5 = 0.

The length of each side of a square is equal to the distance between any two consecutive vertices. Using the distance formula between (-1, 3) and (3, 6), we get side = √((3 - (-1))² + (6 - 3)²) = √(4² + 3²) = √(16 + 9) = √25 = 5 units.

Let the complement angle be x. Then the required angle is x + 40°. Since the sum of complementary angles is 90°, we write x + (x + 40°) = 90°, which gives 2x + 40° = 90° and 2x = 50°, so x = 25°. The other angle is 25° + 40° = 65°. Thus, the angles are 25° and 65°.

The area of a circle is given by A = πr². If the area is an integer multiple of π, then A = Nπ for some integer N. So, πr² = Nπ, which means r² = N, or r = √N. We are looking for the minimum radius r > 0.5. This implies √N > 0.5, so N > 0.25. The smallest integer N that satisfies this condition is N = 1. Therefore, the minimum radius is r = √1 = 1.

In a cyclic quadrilateral ABCD, the sum of opposite interior angles is equal to 180°. Therefore, ∠ABC + ∠ADC = 180°. Given ∠ABC = 70°, we have 70° + ∠ADC = 180°, which gives ∠ADC = 180° - 70° = 110°.

The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circle. Therefore, the angle at the center ∠AOB = 2 × ∠ACB = 2 × 35° = 70°.

Total area of the park = 20 × 15 = 300 m². The area of the inner rectangular region (excluding the path) = Total area - Area of path = 300 - 116 = 184 m². Let the width of the path be x meters. The dimensions of the inner rectangle will be (20 - 2x) and (15 - 2x). Therefore, (20 - 2x)(15 - 2x) = 184. Expanding this gives 300 - 40x - 30x + 4x² = 184, which simplifies to 4x² - 70x + 116 = 0. Dividing by 2, we get 2x² - 35x + 58 = 0. Factoring the quadratic equation gives (2x - 29)(x - 2) = 0. So, x = 2 or x = 29/2 = 14.5. Since the width of the path cannot be 14.5 m (as 2x would be 29 m, which is greater than the park's width of 15 m), the width of the path is x = 2 m.

In a parallelogram, opposite angles are always equal to each other. Therefore, we can set the expressions for the two opposite angles equal to each other: 3x + 10 = 5x - 20. Simplifying this linear equation gives 10 + 20 = 5x - 3x, which means 30 = 2x. Solving for x, we get x = 15.