Question

Example 5 : The shadow of a tower standing
on a level ground is found to be 40 m longer
when the Sun's altitude is 30° than when it is
60°. Find the height of the tower.​

Answer 1

Given :

DB = 40 m.

Sun's altitude = 30° than 60°

According to the question :

Let AB be 'h' 'm' and BC be 'x' 'm'.

DB = 40 m ( given )

So, DB = ( 40 + x ) m.

Two right triangles :

1 ) ΔABC, tan 60° = AB / BC

=> √3 = h / x

2 ) ΔABD, tan 30° = AB / BD

=> 1 / √3 = h / x + 40

Solving ( 1 ) :

1 ) h = x √3

Putting value in ( 2 ) :

( x √3 ) √3 = x + 40

=> 3x = x + 40

=> 3x - x = 40

=> 2x = 40

=> x = 40 / 2

=> x = 20.

So, h = 20 √3

Therefore, The height of the tower is 20 √3.

Answer 2

Question:-

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is 30° than when it is 60°. Find the height of the tower.

Given:-

a) The shadow of a tower standing on a level ground is found to be 40 m longer.

b) The Sun's altitude is 30° than when it is 60°.

Find:- The height of the tower.

Solution:-

According to que.

{ from given & figure }

Let AB be the Height of tower and DC be the length of shadow which is 40 m long and angle ADC is 30° which is Sun's altitude and angle ACB is 60°.

We know that Tower is perperndicular to the ground.{ According to quuestion}

Assume that :-

=> AB = H ... ( 1 )

=> BC = X ... ( 2 )

=> DC = 40 m .... ( 3 )

we know, DC + BC = DB i.e.

=> DB = ( DC + BC )

{ from ( 1 ) & ( 3 ) }

=> DB = ( 40 + X ) ...... ( 4 )

We know there are two triangle's according to figure, Δ ABC &

Δ ABD

we use tan ratio in both triangle to find height of triangle { AB = ? }

Here we also know that value of theta is change in both triangles .

In Δ ABC value of, theta = 30° &

In Δ ABD value of, theta = 60°

In Δ ABC

=> tan(theta) = AB/BC

{ from ( 1 ) & ( 2 ) }

=> tan(60°) = H/X

=> √3 = H/X

=> H = √3 × X ...... ( 5 )

In Δ ABD

=> tan(theta) = AB/DB

=> tan(30°) = H/DB

{ from ( 4 ) }

=> tan(30°) = H/( 40 + X )

=> 1/√3 = H/( 40 + X )

=> √3 × H = 40 + X ...... ( 6 )

From eq. ( 5 ) put value of h in

eq. ( 6 )

=> √3 × √3 × X = 40 + X

=> 3X = 40 + X

=> 3X - X = 40

=> 2X = 40

=> X = 40/2

=> X = 20

Put value of X in eq. ( 5 )

=> H = √3 × X

=> H = √3 × 20 i.e.

=> H = 20√3

Now from eq. ( 1 )

=> AB = H = 20√3 m

Therefore height of the

tower is 20√3 m .

i hope it helps you .