Question
Rohit while seeing a bird on tree top made 45° angle of elevation. He walks 240ft. towards the tree to observe the bird closely, thus making 60° angle of elevation. How far was Rohit from the tree initially?
Answer: Option A
$$\eqalign{
& {\text{Now,}}\,{\text{tan}}{45^ \circ } = 1 = \frac{{PQ}}{{NQ}} \cr
& \therefore {\text{Tree}}\,{\text{height}} \cr
& = PQ = NQ \cr
& = \left( {240 + MQ} \right) \cr
& \tan {60^ \circ } = \sqrt 3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{MQ}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240 + MQ}}{{MQ}} \cr
& \therefore 240 + MQ = \sqrt 3 \,MQ \cr
& \therefore MQ = \frac{{240}}{{\sqrt 3 - 1}}\,{\text{ft}} \cr
& \therefore NQ = 240 + MQ \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240\sqrt 3 }}{{\sqrt 3 - 1}}\,{\text{ft}} \cr} $$
= Rohit was this much far away initially
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$$\eqalign{
& {\text{Now,}}\,{\text{tan}}{45^ \circ } = 1 = \frac{{PQ}}{{NQ}} \cr
& \therefore {\text{Tree}}\,{\text{height}} \cr
& = PQ = NQ \cr
& = \left( {240 + MQ} \right) \cr
& \tan {60^ \circ } = \sqrt 3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{PQ}}{{MQ}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240 + MQ}}{{MQ}} \cr
& \therefore 240 + MQ = \sqrt 3 \,MQ \cr
& \therefore MQ = \frac{{240}}{{\sqrt 3 - 1}}\,{\text{ft}} \cr
& \therefore NQ = 240 + MQ \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{240\sqrt 3 }}{{\sqrt 3 - 1}}\,{\text{ft}} \cr} $$
= Rohit was this much far away initially
Was this answer helpful ?
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