$P$ and $Q$ are two points observed from the top of a building $10\sqrt{3}\ m$ height if the angles of depression of the points are complementary and $PQ=20\ m$, then the distance of $P$ from the building is : (P being the farther point)
Let h = 10*sqrt(3). Angles are θ and 90-θ. Distances are d1 = h/tan(θ) and d2 = h/tan(90-θ) = h*tan(θ). d1 - d2 = 20. h/tan(θ) - h*tan(θ) = 20. 10*sqrt(3) * (1/tan(θ) - tan(θ)) = 20. 1/tan(θ) - tan(θ) = 2/sqrt(3). Let x = tan(θ). 1/x - x = 2/sqrt(3) => (1-x^2)/x = 2/sqrt(3). Solving this leads to tan(θ) = 1/sqrt(3), so θ = 30 degrees. d1 = 10*sqrt(3) / (1/sqrt(3)) = 30.
Let the height of the building be 10 times the square root of 3 meters, and let the angles of depression be theta and 90 degrees minus theta since they are complementary. The horizontal distances from the building to points Q and P are found using basic trigonometry: distance to Q equals height divided by tan(theta), and distance to P equals height divided by tan(90 degrees minus theta), which is height multiplied by tan(theta). This gives the distance to Q as 10 * square root of 3 divided by tan(theta) and the distance to P as 10 * square root of 3 multiplied by tan(theta). Because P is farther than Q, the difference between their distances is the length of PQ, which is 20 meters. We write the equation 10 * square root of 3 * tan(theta) minus 10 * square root of 3 divided by tan(theta) equals 20. Dividing the entire equation by 10 gives square root of 3 * tan(theta) minus square root of 3 divided by tan(theta) equals 2. Letting tan(theta) equal 1 divided by the square root of 3 satisfies this equation, making the distance to P equal to 10 * square root of 3 multiplied by 1 divided by the square root of 3, which is exactly 30 meters.