Answer: his distance from the pole is 90.16 ft
Step-by-step explanation:
based on the diagram of the question i will upload along this answer;
45 - d is equal to part of the pole below the horizontal line
d/x = tan(16) ; x = d/tan(16)............equ1
(45-d) / x = tan(12); x = (45-d) / tan(12)-----------equ2
∴ d/tan(16) = (45-d) / tan(12)
d.tan(12) = (45-d).tan(16)
d.tan12 = (45×tan16) - d.tan16
0.2125d = 12.9034 - 0.2867d
0.2125d + 0.2867d = 12.9034
0.4992d = 12.9034
d = 12.9034/0.4992
d = 25.85 ft
now substitute value of d into any of our previous equation, lets take equation 1
x = 25.85 / tan(16)
x = 25.85 / 0.2867
x = 90.16 ft
Therefore his distance from the pole is 90.16 ft