Question

the shadow of a tower, when the angle of elevation of the sun is 30 is found to be 10 metre longer than when it was 60. the height of the tower is a) 5√3 b) 5(√3-1) c) 5(√3+1) d) 3√5

Answer 1

hey mate here is your answer

Let height of tower CD be h metres

Now in right-angled ΔBCD, we have

x

h

=tan45

o

⇒

x

h

=1

⇒h=x ......(i)

Again in right-angled ΔACD, we have

x+10

h

=tan30

o

⇒

x+10

h

=

3

1

⇒

3

h=x+10

⇒

3

h=h+10 ......using (i)

⇒h(

3

−1)=10

⇒h=

3

−1

10

×

3

+1

3

+1

=

(

3

)

2

−(1)

2

10(

3

+1)

=

3−1

10(1.732+1)

=5×2.732=13.67 m