An electrician wants to repair an electric connection on a pole of height 9m.He needs to reach 1.8m below the top of the pole to do repair work. What should be the length of the ladder which he should use, when he climbs it at an angle 60° with the ground? What will be the distance between foot of the ladder and foot the pole?
Answers
Answer:
The length of the ladder is 8.313 m and he distance between foot of the ladder and foot of the pole is 4.156
Step-by-step explanation:
Length of pole = AC = 9 m
He needs to reach 1.8 m below the top of the pole i.e. AB = 1.8 m
BC = AC-AB = 9-1.8 = 7.2 m
Length of ladder is CD
Now in ΔBCD
Sin \theta = \frac{perpendicular}{hypotenuse}Sinθ=
hypotenuse
perpendicular
Sin 60 = \frac{CB}{CD}Sin60=
CD
CB
\frac{\sqrt{3}}{2} = \frac{7.2}{CD}
2
3
=
CD
7.2
CD = 8.313CD=8.313
Again in ΔBCD
tan \theta = \frac{perpendicular}{Base}tanθ=
Base
perpendicular
tan 60 = \frac{CB}{BC}tan60=
BC
CB
\sqrt{3} = \frac{7.2}{BC}
3
=
BC
7.2
BC= \frac{7.2}{\sqrt{3}}BC=
3
7.2
BC= 4.156BC=4.156
Hence the length of the ladder is 8.313 m and he distance between foot of the ladder and foot of the pole is 4.156
Step-by-step explanation
The length of the ladder is 8.313 m and he distance between foot of the ladder and foot of the pole is 4.156
Step-by-step explanation:
Refer the attached figure
Length of pole = AC = 9 m
He needs to reach 1.8 m below the top of the pole i.e. AB = 1.8 m
BC = AC-AB = 9-1.8 = 7.2 m
Length of ladder is CD
Now in ΔBCD
Sin \theta = \frac{perpendicular}{hypotenuse}
Sin 60 = \frac{CB}{CD}
\frac{\sqrt{3}}{2} = \frac{7.2}{CD}
CD = 8.313
Again in ΔBCD
tan \theta = \frac{perpendicular}{Base}
tan 60 = \frac{CB}{BC}
\sqrt{3} = \frac{7.2}{BC}
BC= \frac{7.2}{\sqrt{3}}
BC= 4.156
Hence the length of the ladder is 8.313 m and he distance between foot of the ladder and foot of the pole is 4.156
Hope this helps you
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