Question

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⁻¹(3√2) and cosec⁻¹(2√2) respectively, then the height of the tower (in metres) is:
(A) 25 (B) 10√5
(C) 100/3√3
(D) 20

Answer 1

The height of the tower is 20m.

Step-by-step explanation:

let the height of the tower be 'h'

         given AB=AC=100m

let BC = a,

∴ $BD=\frac{BC}{2} =\frac{a}{2}$

By geometry AD will be perpendicular to BC, also let D is the point where tower is situated .

∴ From point A

    $=>Cot \theta=\frac{AD}{h}$

   $=>\theta=cot^-(\frac{AD}{h})$

A/Q

    $=>cot^-(\frac{AD}{h} )=cot^-(3\sqrt{2} )$

    $=>\frac{AD}{h} =3\sqrt{2}$

    $=>AD=3\sqrt{2} h$......eq(1)

From point B,

    $=>\theta'=cosec^-(2\sqrt{2} )\\=>cosec\theta'=2\sqrt{2}\\ =>cot\theta'=\sqrt{7}$

      ∴  $\frac{BD}{h} =\sqrt{7}$

      $=>\frac{BD}{h} =\sqrt{7} \\=>BD=h\sqrt{7}$.......eq(2)

NOW , in triangle ABD

  By pythagorous theorom we have

    $AD^2+BD^2=AB^2$

  $=>(3\sqrt{2} h)^2+(\sqrt{7} h)^2=(100)^2$    using eq(1) and eq(2)

  $=>25h^2=(100)^2\\=>h^2=400\\=>h=20$        

Hence the height of the tower is 20 m.