A wire of length 18 m had been tied with electric pole at an angle of elevation $30^{0}$ with the ground.Because it was covering a long distance, it was cut and tied at an angle of elevation $60^{0}$ with the ground. How much length of the wire was cut ?

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Solution

$A D = 18 m$

In $Δ A B D,$

$∠ A B D = 30 °$

$∴ sin 30 ° = \frac{A B}{A D} \Rightarrow A B = A D sin 30 ° = 9 m$

In $Δ A B C,$

$∠ A B C = 90 ° & ∠ A C B = 60 °$

$sin 60 ° = \frac{A B}{A C} = \frac{9}{A C}$

$\frac{\sqrt{3}}{2} = \frac{9}{A C} \Rightarrow A C = \frac{9 \times 2}{\sqrt{3}} = 6 \sqrt{3} m$

$∴$ wire length cut down $= A D - A C = (18 - 6 \sqrt{3}) m = 7.6 m$

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A wire of length 18 m had been tied with electric pole at an angle of elevation 300 with the ground.Because it was covering a long distance, it was cut and tied at an angle of elevation 600 with the ground. How much length of the wire was cut ?

AD=18mIn ΔABD,∠ABD=30°∴sin30°=ABAD⇒AB=ADsin30°=9mIn ΔABC,∠ABC=90°&∠ACB=60°sin60°=ABAC=9AC√32=9AC⇒AC=9×2√3=6√3m∴ wire length cut down =AD−AC=(18−6√3)m=7.6m

A wire of length 18 m had been tied with electric pole at an angle of elevation $30^{0}$ with the ground.Because it was covering a long distance, it was cut and tied at an angle of elevation $60^{0}$ with the ground. How much length of the wire was cut ?