Question 1174478
label your triangle ABCD.
A is the foot of the bottom of the 16 foot wire.
B is the top of the pole
C is the foot of the bottom of the 20 foot wire.
D is the base of the pole.


the height of the pole is BD.
the angle that the second wire makes with the ground is angle BAD.
the angle that the first wire makes with the ground is angle BCD.
the distance between the bottom of the 20 meter wire and the bottom of the 16 meter wire is AC.  AC is the sum of AD and DC.
the length of BC is given as 20 meters.
the length of AB is given as 16 meters.


to find the height of the pole, use sin(48) = BD / 20
solve for BD to get BD = 20 * sin(48) = 14.86289651 meters.


to find the angle that the 16 meter pole makes with the ground, use angle BAD = arcsin(BD/16) = arcsin(14.86289651/16) = 68.26879086 degrees.


to find the length of AD, use 16^2 - BD^2 = 16^2 - 14.86289651^2 = AD^2 and then take the square root of that to find AD = 5.924044847.


to find the length of DC, use 20^2 - BD^2 = 20^2 - 14.86289651^2 = DC^2 and then take the square root of that to find DC = 13.38261213.


those are your answers.


here's a picture to help you see where everything fits.


your solutions are in the picture rounded to 1 d3ecimal place.


<img src = "http://theo.x10hosting.com/2021/020405.jpg" >


i'll be available to answer any questions you night have regarding this.


theo