Verified Solution • 2025–26 Session moderate

QJEEM22Maaa9d: JEE Main 2022 Mathematics - Class 12 JEEM22Maaa9d Solution: A horizontal park is in the shape of a triangle with . A…

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 :

Multiple Choice Options

  • A 323(3−1)\frac{32}{\sqrt{3}}(\sqrt{3}-1)
  • B 323(2−3)\frac{32}{\sqrt{3}}(2-\sqrt{3})
  • C 163(3−1)\frac{16}{\sqrt{3}}(\sqrt{3}-1)
  • D 163(2−3)\frac{16}{\sqrt{3}}(2-\sqrt{3})

Step-by-Step Verified Solution

JEE Main 2022 (Online) 28th July Evening Shift Mathematics - Height and Distance Question 5 English Explanation

OP=OAtan⁡15=OBtan⁡15OP = OA\tan 15 = OB\tan 15 ...... (i)

OP=OCtan⁡45⇒OP=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+13−1)2O{P^2} + 64 = O{P^2}{\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right)^2}

64=OP2[(3+1)2−(3−1)2(3−1)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(3−1)2)= O{P^2}\left( {{{4\sqrt 3 } \over {{{\left( {\sqrt 3 - 1} \right)}^2}}}} \right)

OP2=64(3−1)243=323(2−3)O{P^2} = {{64{{\left( {\sqrt 3 - 1} \right)}^2}} \over {4\sqrt 3 }} = {{32} \over {\sqrt 3 }}\left( {2 - \sqrt 3 } \right)

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring