A B C is a triangular park with A B = A C = 100 metres. A vertical tower is situated at the mid-point of BC.…

ABC is a triangular park with AB=AC=100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at, A and B are cot-132 and cosec-122 respectively, then the height of the tower (in metres) is
  1. 10033
  2. 20
  3. 25
  4. 105

Solution

Let AD=y & DE=h (Tower)

∴ y2=1002-x2

Now cotθ=32

⇒tanθ=132=hy

⇒y=32h

⇒y2=1002-x2=18h2  …1

⇒tan∅=17=hx

⇒x=h7    …2

Form 1 & 2 1002-7h2=18h2

⇒ 25h2=1002⇒h=20

Asked in: JEE Main 2019 (10 Apr Shift 1)

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