Question

The length of the shadow of a tower standing on level plane is found to be 2x metres longer when the sun’s altitude is 30° than when it was 45°. Prove that the height of tower is metres.

Answer 1

Answer:

The height of tower is 2.7 x meters

Step-by-step explanation:

Given as :

The height of tower = OC = h meters

The measure of distance of point A from O = OA = y meters

The measure of distance of point B from O is 2 x meter longer = OB = ( y + 2 x) meters

According to question

From figure

In Triangle OAC

Tan angle = $\dfrac{perpendicular}{base}$

i.e Tan 45° = $\dfrac{OC}{OA}$

Or,  1 = $\dfrac{h}{y}$

∴  y = h                    ..........1

Again

In Triangle OBC

Tan angle = $\dfrac{perpendicular}{base}$

i.e Tan 30° = $\dfrac{OC}{OB}$

Or,  $\dfrac{1}{\sqrt{3} }$ = $\dfrac{h}{2x + y}$

∴         2 x + y = √3 h           ..........2

from eq 1 and eq 2

Put the value of y

2 x + h = √3 h  

Or, 2 x =√3 h - h

Or, 2 x = h ( √3 - 1 )

∴ h = $\dfrac{2x}{\sqrt{3} -1}$

Or, h = 2.7 x  meter

So, The height of tower = h = 2.7 x meters

Hence,  The height of tower is 2.7 x meters    Answer

Answer 2

$\huge{ \mathfrak{ \overline{ \underline{ \underline{ \yellow{answer☺︎︎}}}}}}...</p><p> </p><p></p><p></p><p>[tex]\huge\color{yellow}{\underline{\underline{ɾεƒεɾร \: ƭσ \: ƭɦε \: αƭƭαcɦɱεɳƭ \: \::}}}$