Question

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 1

Answer:

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Answer 2

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