Distance between two poles is 65m 20cm in between the poles there is a tree which is 32m 45cm away from the first pole find the distance between the tree and the second pole
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let AB be the 1st pole and CD be the 2nd pole
AC=15m
CD=24m
let AB=x
CE=x
ED=24-x
consider triangle BED
tan30° = DE/BE
1/√3 = (24-x)/15
15= √3(24-x)
5√3 = 24-x
x = 24-5√3
= 15.33m
AC=15m
CD=24m
let AB=x
CE=x
ED=24-x
consider triangle BED
tan30° = DE/BE
1/√3 = (24-x)/15
15= √3(24-x)
5√3 = 24-x
x = 24-5√3
= 15.33m
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