Question

a circus artist is climbing from the ground along a rope stressed from the top of a vertical rope and tide at the ground the height of the pole is 12 metre and the angle made by the rope with the ground level is 30 calculate the distance covered by the artist in climbing to the top of phone and also find the distance between bottom of the poles and ground.Of the rope​

Answer 1

Question:

A circus artist is climbing from the ground along a rope stressed from the top of a vertical rope and tide at the ground the height of the pole is 12 metre and the angle made by the rope with the ground level is 30 calculate the distance covered by the artist in climbing to the top of phone and also find the distance between bottom of the poles and ground of the rope.

Answer:

let AB be the height of the pole which is 12m,HC be the ground level ,AC be the rope tide at the ground and make a elevation angle ACB with the ground level which is 30°.We need to find the height of rope and distance between the bottom of pole and ground point of rope.

To find the height of the rope :

p/h= sinA

→12/h= sin30°

→12/h=1/2

→h=24

To find the distance between pole and ground point of rope .

p/b=tanA

→p/30=tan30°

→p/30=1/√3

→√30p=30

→p=30/√3

Rationalising the denominator

→p30/√3=√3/√3

→p=30√30/3

→P= 10/√30

Hence the height of rope = 24m and the distance between the foot of pole and ground point of rope =10/√3cm

Answer 2

GiveN :

• Height of Pole (BC) = 12 m
• Angle made by Rope (Ø) = 30°

To FinD :

• Distance covered by the Artist (AC)
• Distance between bottom of Pole and ground (AB)

SolutioN :

1) First we will find the Distance covered by the Artist (AC)

$\underbrace{\sf{In \: \triangle \: ABC}}$

$\implies \sf{\sin \theta \: = \: \dfrac{Perpendicular}{Hypotenuse}} \\ \\ \implies \sf{\sin \theta \: = \: \dfrac{BC}{AC}} \\ \\ \implies \sf{\sin 30^{\circ} \: = \: \dfrac{12}{AC}} \\ \\ \implies \sf{\dfrac{1}{2} \: = \: \dfrac{12}{AC}} \\ \\ \implies \sf{AC \: = \: \dfrac{12 \: \times \: 2}{1}} \\ \\ \implies \sf{AC \: = \: 24 \: cm} \\ \\ \underline{\sf{\therefore \: Distance \: traveled \: by \: Artist \: is \: 24 \: m}}$

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2) Distance between the bottom of the pole and ground

$\implies \sf{\tan \theta \: = \: \dfrac{Perpendicular}{Base}} \\ \\ \implies \sf{\tan 30^{\circ} \: = \: \dfrac{AB}{12}} \\ \\ \implies \sf{\dfrac{1}{\sqrt{3}} \: = \: \dfrac{AB}{12}} \\ \\ \implies \sf{AB \: = \: \dfrac{12}{\sqrt{3}}} \\ \\ \implies \sf{AB \: = \: \dfrac{12}{\sqrt{3}} \: \times \: \dfrac{\sqrt{3}}{\sqrt{3}}} \\ \\ \implies \sf{AB \: = \: \dfrac{12 \: \times \: \sqrt{3}}{3}} \\ \\ \implies \sf{AB \: = \: 4 \sqrt{3}} \\ \\ \underline{\sf{\therefore \: Distance \: between \: bottom \: of \: pole \: and \: ground \: is \: 4 \sqrt{3} \: m}}$