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Height and Distance
Height and Distance
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Question

A horizontal park is in the shape of a triangle OAB\mathrm{OAB} with AB=16\mathrm{AB}=16. A vertical lamp post OP\mathrm{OP} is erected at the point O\mathrm{O} such that PAO=PBO=15\angle \mathrm{PAO}=\angle \mathrm{PBO}=15^{\circ} and PCO=45\angle \mathrm{PCO}=45^{\circ}, where C\mathrm{C} is the midpoint of AB\mathrm{AB}. Then (OP)2(\mathrm{OP})^{2} is equal to :

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Solution

OP=OAtan15=OBtan15OP = OA\tan 15 = OB\tan 15 ...... (i) OP=OCtan45OP=OCOP = OC\tan 45 \Rightarrow OP = OC ...... (ii) OA=OBOA = OB ....... (iii) OC2+82=OA2O{C^2} + {8^2} = O{A^2} OP2+64=OP2(3+131)2O{P^2} + 64 = O{P^2}{\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right)^2} 64=OP2[(3+1)2(31)2(31)2]64 = O{P^2}\left[ {{{{{\left( {\sqrt 3 + 1} \right)}^2} - {{\left( {\sqrt {3 - 1} } \right)}^2}} \over {{{\left( {\sqrt 3 - 1} \right)}^2}}}} \right] =OP2(43(31)2) = O{P^2}\left( {{{4\sqrt 3 } \over {{{\left( {\sqrt 3 - 1} \right)}^2}}}} \right) OP2=64(31)243=323(23)O{P^2} = {{64{{\left( {\sqrt 3 - 1} \right)}^2}} \over {4\sqrt 3 }} = {{32} \over {\sqrt 3 }}\left( {2 - \sqrt 3 } \right)

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