Question
Tree top’s angle of elevation is 30° from a point on ground, 300m away the tree. When the tree grew up its angle of elevation became 60° from the same point. How much did the tree grow?
Answer: Option B
$$\eqalign{
& {\text{Original tree height}} = {\text{h}} = {\text{MQ}} \cr
& {\text{New}}\,{\text{tree}}\,{\text{height}} = PQ \cr
& {\text{in}}\,\Delta MQN,\,\tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{NQ}} \cr
& \therefore MQ = \frac{{300}}{{\sqrt 3 }} \cr
& {\text{in}}\,\Delta PQN,\,\tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{{NQ}} \cr
& \therefore PQ = 300\sqrt 3 \cr
& {\text{Tree}}\,{\text{grew}} = PQ - MQ \cr
& \therefore {\text{Tree}}\,{\text{grew}} \cr
& = 300\sqrt 3 - \frac{{300}}{{\sqrt 3 }} \cr
& = 300\frac{2}{{\sqrt 3 }} \cr
& = 3 \times 100 \times \frac{2}{{\sqrt 3 }} \cr
& = 200\sqrt 3 \,{\text{m}} \cr} $$
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$$\eqalign{
& {\text{Original tree height}} = {\text{h}} = {\text{MQ}} \cr
& {\text{New}}\,{\text{tree}}\,{\text{height}} = PQ \cr
& {\text{in}}\,\Delta MQN,\,\tan {30^ \circ } = \frac{1}{{\sqrt 3 }} = \frac{{MQ}}{{NQ}} \cr
& \therefore MQ = \frac{{300}}{{\sqrt 3 }} \cr
& {\text{in}}\,\Delta PQN,\,\tan {60^ \circ } = \sqrt 3 = \frac{{PQ}}{{NQ}} \cr
& \therefore PQ = 300\sqrt 3 \cr
& {\text{Tree}}\,{\text{grew}} = PQ - MQ \cr
& \therefore {\text{Tree}}\,{\text{grew}} \cr
& = 300\sqrt 3 - \frac{{300}}{{\sqrt 3 }} \cr
& = 300\frac{2}{{\sqrt 3 }} \cr
& = 3 \times 100 \times \frac{2}{{\sqrt 3 }} \cr
& = 200\sqrt 3 \,{\text{m}} \cr} $$
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