a) $$10\sqrt 3 \,{\text{m}}$$
b) $$\frac{{10}}{{\sqrt 3 }}\,{\text{m}}$$
c) $$\frac{{\left( {\sqrt 2 – 1} \right)}}{{10}}\,{\text{m}}$$
d) $$\frac{{10}}{{\sqrt 2 }}\,{\text{m}}$$
Answer: a
Explanation:
$$\eqalign{
& {\text{in}}\,\Delta MNQ,\tan {30^ \circ } = \frac{{MQ}}{{NQ}} \cr
& \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{10}} \cr
& MQ = \frac{{10}}{{\sqrt 3 }} \cr
& {\text{Also}}\,{\text{by}}\,{\text{Pythagoras}}\,{\text{theorem}} \cr
& M{N^2} = M{Q^2} + N{Q^2} \cr
& {L^2} = \frac{{100}}{3} + 100 \cr
& L = \frac{{20}}{{\sqrt 3 }} \cr
& {\text{Height}}\,{\text{of}}\,{\text{tree}} = L + MQ \cr
& = \frac{{20}}{{\sqrt 3 }} + \frac{{10}}{{\sqrt 3 }} \cr
& = \frac{{30}}{{\sqrt 3 }} \cr
& = \frac{{3 \times 10}}{{\sqrt 3 }} \cr
& = 10\sqrt 3 \,{\text{m}} \cr} $$
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