Question

An electric pole is 12 m high . A steel wire tied to top of the pole is affixed at a point on the ground to keep the pole upright . If the wire makes an angle of 60 ° with the horizontal through the foot of the pole , find the length of the wire .​

Answer 1

Let AB=10= Height of Pole

Let AB=10= Height of Poleand AD be the length of the wire

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABD

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o =

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = AD

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 =

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10 ⇒AD=10

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10 ⇒AD=10 2

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10 ⇒AD=10 2

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10 ⇒AD=10 2

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10 ⇒AD=10 2 ⇒10×1.414 (Take

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10 ⇒AD=10 2 ⇒10×1.414 (Take 2

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10 ⇒AD=10 2 ⇒10×1.414 (Take 2

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10 ⇒AD=10 2 ⇒10×1.414 (Take 2 =1.414)

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10 ⇒AD=10 2 ⇒10×1.414 (Take 2 =1.414) =14.14 m

Let AB=10= Height of Poleand AD be the length of the wireFrom △ABDsin45 o = ADAB ⇒ 2 1 = AD10 ⇒AD=10 2 ⇒10×1.414 (Take 2 =1.414) =14.14 m