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?
Answer & explanation
Correct answer: option 1
⇒ 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.