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Q2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30º with it. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree.
Answer :


In \( ∆ \ ABC \)



\( \frac{AB}{BC} \ = \ tan30º \ => \ \frac{AB}{8} \ = \ \frac{1}{ \sqrt{3}} \)

=> \( AB \ = \ \frac{8}{ \sqrt{3}} \)      (1)

And, \( \frac{AC}{BC} \ = \ sec30º \ => \ \frac{AC}{8} \ = \ \frac{2}{ \sqrt{3}} \)

=> \( AC \ = \ \frac{16}{ \sqrt{3}} \)      (2)

Now, Height of the tree = AB + AC

\( =\ \frac{8}{ \sqrt{3}} \ + \ \frac{16}{ \sqrt{3}} \ = \ \frac{24}{ \sqrt{3}} \)

\( = \ \frac{24}{ \sqrt{3}} \ × \ \frac{ \sqrt{3}}{ \sqrt{3}} \ = \ 8\sqrt{3} \) m