The height of tower is 30 m, find the length of the shadow of the tower, when the angle of elevation of the sun is 45°.
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Answer:
Let AB be the tower and let AC and AD be its shadows when the angles of elevation of the sum are 60
o
and 45
o
, respectively. Then,
∠ACB=60
o
,∠ADB=45
o
,CD=10 m
Let AB=h metres and AC=x metres
In right △CAB, we have,
tan60
o
=
AC
AB
3
=
x
h
x=
3
h
......(1)
In right △ DAB, we have,
tan45
o
=
AD
AB
1=
10+x
h
x=h−10 .....(2)
From 1 and 2, we get,
3
h
=h−10
h=
3
h−10
3
3
h−h=10
3
h(
3
−1)=10
3
h=5
3
(
3
+1)
h=15+5
3
m
solution
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