The length of the shadow of a vertical pole on the ground is 18 m.If the angle of elevation of the sun at that time is $\theta$, such that $\cos\theta=\frac{12}{13}$, then what is the height (in m)of the pole?

7.5

⇒ cos \(\theta \) = \(\frac{12}{13}\)

⇒ 12 units = 18m

⇒ 1 unit = \(\frac{18}{12}\) = \(\frac{3}{2}\)

⇒ 13 units = \(\frac{3}{2}\) x 13 = 19.5m

⇒ AC = 19.5m

Now, using pythagoras theorem,

⇒ \( {19.5 }^{ 2} \) = \( {18 }^{ 2} \) + \( {AB }^{ 2} \)

⇒ \( {AB }^{ 2} \) = 380.25 - 324

⇒ \( {AB }^{ 2} \) = 56.25

⇒ AB = 7.5m

Therefore, height of the pole is 7.5m.