Question

A pole stands in a park such that is shadow increases by 2 m when angle of elevation changes from 45 to 30 then find the height​

Answer 1

..............................

Answer 2

Answer:

The height of the pole is approximately 2.73m

Step-by-step explanation:

Find the attachment below for reference.

Let the pole(AB) be of height h

When the angle of elevation is 45°, let the shadow be formed at C

Let the length of this shadow(AC) be x m

When the angle of elevation is 30°, let the shadow be formed at D

As the shadow increases by 2m

Let the length of this shadow(AD) will be x+2 m

We know that tan $\theta$ = $\frac{Perpendicular}{Base}$

Therefore in Δ ABC,

tan 45° =  $\frac{h}{x}$

=> 1 = $\frac{h}{x}$               (as tan 45° = 1)

=> x = h                                 --(i)

in Δ ABD,

tan 30° =  $\frac{h}{x+2}$

=> $\frac{1}{\sqrt{3} }$ = $\frac{h}{x+2}$              (as tan 45° = 1)

=> x+2 = h √3

=> h + 2 = h √3               (from equation(i))

=> 2 = h √3 - h

=> 2 = h( √3 - 1)

=> h = $\frac{2}{\sqrt{3} -1 }$

We will now rationalize the denominator.

=> h = $\frac{2}{\sqrt{3} -1 }$ * $\frac{\sqrt{3} +1}{\sqrt{3} +1 }$

=> h =  $\frac{2(\sqrt{3} +1)}{(\sqrt{3} +1 )(\sqrt{3} +1) }$

=> h =  $\frac{2(\sqrt{3} +1)}{(\sqrt{3})^{2} -1 ^{2} }$

=> h =   $\frac{2(\sqrt{3} +1)}{3-1}$

=> h =   $\frac{2(\sqrt{3} +1)}{2 }$

=> h = √3 + 1

=> h = 1.732 + 1

=> h = 2.732m

Therefore, the height of the pole is approximately 2.73m