Answer: the distance between him and his friend is 218.39 feet
Step-by-step explanation:
Given the data in the question;
FROM IMAGE A;
distance travelled in minute and a half
⇒ (60 + 30)sec × 4ft/sec = 360 fts
FROM IMAGE B
tan35°= h/x --- equ 1
and tan36° = h/(360 - x), so h = (360 - x)tan36°
we substitute value of h into euq 1
tan35° = (360 - x)tan36°/x
xtan35° = (360 - x)tan36°
0.7002x = 261.5553 - 0.7265x
0.7002x + 0.7265x = 261.5553
1.4267x = 261.5553
x = 183.32 feet
so
360 - x ⇒ ( 360 - 183.32) = 176.68
FROM IMAGE C
Let the distance between them be d
so
cos36° = 176.68 / d
dcos36° = 176.68
d = 176.68 / 0.809
d = 218.39 feet
Therefore the distance between him and his friend is 218.39 feet