Two vertical poles of lengths 11 ft and 14 ft stand 16 ft apart. A cable reaches from the top of one pole to some point on the ground between the poles and then to the top of the other pole.
Let the distance of that point from smaller pole( 11 ft) be x ft. Hence the distance of that point from the larger pole would be (16-x)ft.
The cable forms a right angle triangle with each pole. In Right angle triangle Hypotenuse^2 = Base^2 + Perpendicular^2
The length of the cable connecting the point and the smaller pole is √(11² + x²)
The length of the cable connecting the point and the larger pole is √(14² + (16-x)² )
Total length of the cable is 30 ft
So, √(11² + x²) + √(14² + (16-x)² ) = 30
or, √(11² + x²) - 30 = √(14² + (16-x)² )
squaring both sides,
(11² + x²) + 900 - 60√(11² + x²) = (14² + (16-x)² )
or, 121 + x² + 900 - 60√(11² + x²) = 196 + 256 + x² - 32x
or, 32x + 569 = 60√(11² + x²)
Squaring both sides,
or, (32x + 569)² = 3600(11² + x²)
or, 1024x² + 323761 + 36416x = 3600( 121 + x²)
or, 1024x² + 323761 + 36416x = 435600 + 3600x²
or, 2576x² -36416x + 111839 = 0
Using the quadratic formula, x = 9.6267 ft or 4.5099 ft
Therefore, the point is located at a distance of 9.6267 ft or 4.5099 ft from the smaller pole of height 11 ft.
Hope this helps ..!!
Thank you :)