A wire of length 18 m had been tied with electric pole at an angle of elevation 30° with
the ground. Because it was covering a long distance, it was cut and tied at an angle of
elevation 60° with the ground. How much length of the wire was cut?
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Answer:
AD=18m
In ΔABD,
∠ABD=30°
∴sin30°=
AD
AB
⇒AB=ADsin30°=9m
In ΔABC,
∠ABC=90°&∠ACB=60°
sin60°=
AC
AB
=
AC
9
2
3
=
AC
9
⇒AC=
3
9×2
=6
3
m
∴ wire length cut down =AD−AC=(18−6
3
)m=7.6m
solution
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