Question

An electrician has to repair an electric fault on a pole of height 4m.She needs to reach a point 1.3m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach
the required position? Also how far from the foot of the pole should she place the foot of the ladder? ( You may take √3 = 1.73)

Answer 1

this answer is given by applying the value of √3 = 1.732

Answer 2

Answer:

i)Length from foot of the pole to foot of the ladder =1.557 m

=1.557 mii)Length of the ladder = 3.114 m

Step-by-step explanation:

Height of the electric pole (BD)=4 m

Height of the repairing place below the top of the pole (CD)=1.3 m

BC = BD - CD = 4-1.3 = 2.7 m

Let length of the ladder = y m

Length from foot of the pole to foot of the ladder (AB)= x m

<CAB = 60° , <B = 90°

$i) In \: \triangle ABC ,\\tan60\degree = \frac{BC}{AB}$

$\implies \sqrt{3}=\frac{2.7}{x}$

$\implies x = \frac{2.7}{\sqrt{3}}$

$\implies x = \frac{2.7\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$

$\implies x = \frac{2.7\times 1.73}{3}$

$\implies x = 0.9 \times 1.73\\=1.557 \:m$

$ii) In \: \triangle ABC ,\\sin60\degree = \frac{BC}{AC}$

$\implies \frac{\sqrt{3}}{2}=\frac{2.7}{y}$

$\implies y= \frac{2.7\times 2}{\sqrt{3}}$

$\implies y = \frac{2.7\times2\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$

$\implies y = \frac{2.7\times \times 1.73}{3}$

$\implies y = 0.9\times 2 \times 1.73\\=3.114 \:m$

Therefore,

i)Length from foot of the pole to foot of the ladder (AB)= x m=1.557 m

ii)Length of the ladder = y m= 3.114 m

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