the horizontal distance between two poles is15m.the angle of depression of the top of the first pole as seen from the top of the second pole is 30degree.if the hight of the second pole is 24m,find the hight of the first 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
Hope u like it
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
Hope u like it
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