Question

A tower is 50m high . its shadow is 'x'metres shorter when the sun's altitude is 45 degree than when it is 30 degree .find the value of 'x'

Answer 1

height of tower =50m
angle=45
length of shadow= 'x' m

tan45=opp.÷adj.
1=opp.÷adj.(since tan45 is 1)
opp.=adj.
length of shadow =50m

Answer 2

36.6 m is the length of the shadow.  

Given:

The height of the tower is 50 m.

Its shadow is 'x' metres shorter when the sun's altitude is 45 degree than when it is 30 degree

To find: x

Solution:

Let AB be the height of the tower = 50m

C be the point at 45° and D be at 30° of the sun’s altitude.

$\begin{array}{l}{\tan 45^{\circ}=\frac{\text {height of the tower }}{\text {length of the shadow}}=\frac{50}{\text {length of the shadow}}} \\ {1=\frac{50}{\text {length of the shadow}}}\end{array}$

Length of the shadow at 45ᵒ = 50m

$\begin{array}{l}{\tan 30^{\circ}=\frac{\text {height of the tower }}{\text {length of the shadow}}=\frac{50}{\text {length of the shadow}}} \\ {\frac{1}{\sqrt{3}}=\frac{50}{\text {length of the shadow}}}\end{array}$

Length of the shadow at 30ᵒ $= 50\sqrt{3} m$

The distance between 45ᵒ and 30ᵒ

$= \begin{array}{l}{=50 \sqrt{3}-50} \\ {50(\sqrt{3}-1)=50(0.732)} \\ {=36.6 \mathrm{m}}\end{array}$

$x = 36.6 m$

Length of the shadow $\bold{= 36.6 m.}$