Verified Solution • 2025–26 Session moderate

QJEEM22Ma4b6f: JEE Main 2022 Mathematics - Class 12 JEEM22Ma4b6f Solution: A tower PQ stands on a horizontal ground with base on the…

A tower PQ stands on a horizontal ground with base QQ on the ground. The point RR divides the tower in two parts such that QR=15 mQ R=15 \mathrm{~m} . If from a point AA on the ground the angle of elevation of RR is 60∘60^{\circ} and the part PRP R of the tower subtends an angle of 15∘15^{\circ} at AA , then the height of the tower is :

Multiple Choice Options

  • A 5(23+3) m5(2 \sqrt{3}+3) \,\mathrm{m}
  • B 5(3+3) m5(\sqrt{3}+3) \,\mathrm{m}
  • C 10(3+1) m10(\sqrt{3}+1) \,\mathrm{m}
  • D 10(23+1) m10(2 \sqrt{3}+1) \,\mathrm{m}

Step-by-Step Verified Solution

JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Height and Distance Question 8 English Explanation

For Δ\Delta AQR,

tan⁡60∘=15x\tan 60^\circ = {{15} \over x} ....... (1)

From Δ\Delta AQP,

tan⁡75∘=hx\tan 75^\circ = {h \over x}

⇒(2+3)=hx\Rightarrow \left( {2 + \sqrt 3 } \right) = {h \over x} [ ∵\because tan⁡75∘=2+3\tan 75^\circ = 2 + \sqrt 3 ]

⇒h=(2+3)x\Rightarrow h = \left( {2 + \sqrt 3 } \right)x

=(2+3)153= \left( {2 + \sqrt 3 } \right){{15} \over {\sqrt 3 }} [From (1)]

=(2+3)×1533= \left( {2 + \sqrt 3 } \right) \times {{15\sqrt 3 } \over 3}

=(2+3)×53= \left( {2 + \sqrt 3 } \right) \times 5\sqrt 3

=5(23+3)= 5\left( {2\sqrt 3 + 3} \right) m

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