Question

) A tower stands on a horizontal plane. The
shadow of the tower when the angle of
the elevation of the Sun is 30° is 45 m
more than when the angle of elevation of
the Sun is 60°, Then, the height of the
tower is ...
m.​

Answer 1

Given :-

• A tower stands on a horizontal plane

• The shadow of the tower when the angle of elevation of the sun is 30° is 45m more than when the angle of elevation of the sun is 60°

Solution :-

Let the length BC be xm and height of the tower be hm

Therefore,

The length of shadow that is BD

= ( x + 45)m

Now, In ΔABC

tan60° = AB/BC

Subsitute the required values,

√3 = h/x

[ The value of tan60° = √3]

√3x = h

x = h/√3. ...eq( 1 )

Now, In ΔABD

tan45° = AB/BD

Subsitute the required values,

1 = h / ( x + 45)

x + 45 = h. ...eq( 2 )

Subsitute eq( 1 ) in eq( 2 )

h/√3 + 45 = h

h + 45√3 / √3 = h

h + 45√3 = √3h

45√3 = √3h - h

45√3 = h( √3 - 1 )

45√3/ (√3 - 1 ) = h

By rationalising the denominators we get :-

= 45√3 * ( √3 + 1 ) / ( √3 - 1 ) * ( √3 + 1 )

= 45 * 3 + 45√3 / 3 - 1

= 45( 3 + √3)/2

= 45 ( 3 + 1.73)/2

= 45 * 4.73/2

= 212.85/2

= 106.42

Hence, The height of the tower is 106.42m^2 $.$