Question

The horizontal distance between two poles is 15m .the angle of depression of the top of first pole as seen from the top of the second pole is 30degree .if the height of the second pole is 24m.find the height of the first pole

Answer 1

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

Answer 2

Let AB be the pole of height h metres and CD be the second pole of height 24m.

Draw CF||BD and AE ||BD.Then,

BD = AE = 15., CE = (24 - h) m

$\sf\angle{{CAE}=\angle{ACF}=30°}$

From right ∆AEC, we have

$\longrightarrow$$\sf{tan\:30°= \frac{CE}{AE}=\frac{1}{ \sqrt{3} } = \frac{24 - h}{15}}$

$\longrightarrow$$\sf{\frac{15}{ \sqrt{3} } = 24 - h}$

$\longrightarrow$$\sf{h = 24 - \frac{15}{ \sqrt{3} } = 24 - 5 \sqrt{3}}$

$\longrightarrow$$\sf{24-5×1.732}$

$\longrightarrow$$\sf{24-8.66}$

$\longrightarrow$${\boxed{\sf{\red{15.34m}}}}$

Hence, the height of the first pole = 15.34m.