Question

A wire of length 18 m had been tied with electric pole at an angle of elevation 30º with the ground. Because it was covering a long distance, it was cut and tied at an angle of elevation 60º with the ground. How much length of the wire was cut?

Answer 1

Answer:

sin 30° = P/H

1/2 = P/18

9m= P

when , angle is 60°

sin 60°= 9/H

√3/2 = 9/H

18/√3 = H

multiply and divide by √3

so , H = 18√3/3

hence ,H =6√3 m

Answer 2

Answer:

7.60m

Step-by-step explanation:

Length of the wire AC = 18m  (Given)

Wire angle of elevation with the ground = 30°  (Given)

Wire was cut and tied at an angle of elevation = 60°  (Given)

Length of the remaining wire after cutting = AD

Let the height of electric pole AB = h

In right ΔACB

Sin 30 = AB/AC = h/18

1/2 = h/18

h = 9

In right ΔADB

Sin 60 = AB/AD = h/AD

√3/2 = h/AD

AD = 6√3 ( after simplifying)

Length of the wire was cut = AC - AD

= 18 - 6√3

= 18 - 6×1.732

= 18 - 10.39

= 7.60

Thus, the wire was cut 7.60m