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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