An engineer standing at a point P wishes to determine the width of a new rectangular pond. She finds the distance to the western-most point A of the pond from P to be 60m, while the distance to the northern-most point B of the pond from P is 80m. If the angle between the two lines of sight at P is 60°, then the width AB (in metres) of the pond is (AB is not parallel to line of North-South)
We can model this problem using a triangle formed by points P, A, and B. Given parameters: - The distance \( PA = 60 \text{ m} \) - The distance \( PB = 80 \text{ m} \) - The angle between the lines of sight at P is \( \angle APB = 60^\circ \). We need to find the length of side AB. This can be calculated using the Law of Cosines: \[ AB^2 = PA^2 + PB^2 - 2(PA)(PB)\cos(\angle APB) \] Substitute the given values into the formula: \[ AB^2 = 60^2 + 80^2 - 2(60)(80)\cos(60^\circ) \] We know that \( \cos(60^\circ) = \frac{1}{2} \). \[ AB^2 = 3600 + 6400 - 2(4800)\left(\frac{1}{2}\right) \] \[ AB^2 = 10000 - 4800 \] \[ AB^2 = 5200 \] Now, take the square root to find AB: \[ AB = \sqrt{5200} = \sqrt{400 \times 13} \] \[ AB = 20\sqrt{13} \text{ m} \] Therefore, the width AB of the pond is \( 20\sqrt{13} \) metres.
Reinforce your conceptual grasp with related questions from NIMCET 2026:
Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.
Start Free Mock Test Series →