Question

helgULUI IJU V
10.
Clinky observes a tower PQ of height 'h' from a point A on the ground. She moves a
distance d towards the foot of the tower and finds that the angle of elevation has
direction and finals that the angle of elevation is 3 times at A. Prove that 36 =35
ing with the wind in a horizontal line at a height of​

Answer 1

Let RB = x

In ? PBQ, angle BQR is exterior angle

∴ ∠PBQ = 2α – α = α

Now in ? PBQ, ∠ PBQ = ∠QPB

This gives PQ = QB = d

Similarly, ∠BRA is exterior angle of ? BQR

∴ ∠QBR = 3α - 2α = α

And ∠BRQ = π – 3 α (linear pair)

Now in ?BQR, by applying Sine rule, we get

d/sin (π - 3α) = 3d/4 / sin α = x/sin 2α

⇒ d/sin 3α = 3d/4 sin α = x/sin2 α

⇒ d/3 sin α – 4 sin3α = 3d/4 sin α = x/2sin α cos α

⇒ d/3 – 4 sin2 α = 3d/4 = x/2cos α . . . . . . . . . . . . . . . . . . . . . . (A)

From eq. (A), taking first two parts

⇒ d/3 – 4 sin2 α = 3d/4 ⇒ 4 = 9 – 12 sin2 α

⇒ sin2 α = 5/12 ⇒ cos2 α = 7/12

Also from eq. (A) using last two parts, we have

3d/4 = x/2 cos α ⇒ 4x2 = 9 d2 cos2 α

x2 = 9d2/4 = 7/12 = 21/16 d2 . . . . . . . . . . . . . . . . . . . . (B)

Again from ? ABR, we have sin 3α = h/x

⇒ 3 sin α – 4 sin3 α = h/x ⇒ sin α(3 – 4 sin2 α) = h/x

⇒ sin α [3 – 4 x 5/12] = h/x (using sin2 α = 5/12)

⇒ 4/3 sin α = h/x

Squaring both sides, we get

16/9 sin2 α = h2/x2 16/9 . 2/12 = h2/x2

(again using sin2 α = 5/12)

⇒ h2 = 4 . 5/9 . 3x2

⇒ h2 = 20/27 . 21/16 d2

[using value of x2 from eq. (B)]

⇒ h2 = 35/36 d2 ⇒ 36 h2 = 35 d2

Hence, 36 h2 = 35 d2.

Answer 2

Explanation:

let the angle of elevation at A be anglePAQ = a,then

anglePBQ =2a, anglePCQ =3a and AB=d;BC=3d/4

apply sine rule to PBC, sin(180-3a)/d = sina/(3d/4)

3/4 = sina/sin3a

3/4 = 1/(3-4sin^2a)

9-12sin^a = 4

sin^2a = 5/12

cos^2a = 7/12

from trianglePBQ,

sin2a=h/d

4sin^2a.cos^2a = h^2/d^2

4x(5/12)x(7/12)=h^2/d^2

36h^2 = 35d^2