Question

A man wishes to estimate the distance of a nearby tower from him. He stands at a point A in front of the tower C and spots a very distant object O in line with AC. He then walks perpendicular to AC up to B, a distance of 100m, and looks at O and C again. Since O is very distant, the direction BO is practically the same as AO; but he finds the line of sight of C shifted from the original line of sight by an angle ϴ= 40⁰ (ϴ is known as ‘parallax) estimate the distance of the tower C from his original position A.​

Answer 1

$\underline{\underline{\red{\textbf{Step - By - Step - Explanation : -}}}}$

To Find :

• we have to find the distance from position A to C.

Solution :

• parallax angle θ = 40°
• AB = 100 m

Let r be the required distance and l be the distance between A and B.

we know that,

$\star\boxed{\underline{\blue{\bf{l= r\theta}}}}$

here,

• l = 100m
• θ = 40°
• r = ???

Changing theta into radian

⠀⠀⠀⠀⠀40 × π/180

⠀⠀⠀⠀⠀2π/9 radian

Now,

⠀⠀⠀›› 100 = r × 2π/9

⠀⠀⠀›› 100 = r × (2×22/7)/9

⠀⠀⠀›› 100 = r × (44/7)/9

⠀⠀⠀›› 100 = r × 44/7 × 1/9

⠀⠀⠀›› 100 = r × 44/63

⠀⠀⠀›› 100 × 64/44 = r

⠀⠀⠀›› r = 6300/44

⠀⠀⠀›› r = 143 m (approx)

Hence

• Distance of tower from position A to C is 143m.

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