Question

 If a kite is flying at a height of 40√3m from the level ground, attached to a string inclined at 60 degree to the horizontal, then the length of the string is​

Answer 1

ANSWER:

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Here,

$\;\;\;\bullet\;\;$ Height of kite from ground is 40√3 m.

$\;\;\;\bullet\;\;$ Let Height of string attached with kite be x.

$\;\;\;\bullet\;\;$ Angle of elevation = 60°

★ In ∆ABC,

$\star\;\sf sin\;60^\circ = \dfrac{BC}{AC} = \dfrac{P}{H}$

$:\implies\sf \dfrac{ \sqrt{3}}{2} = \dfrac{40\sqrt{3}}{x}$

$:\implies\sf \sqrt{3} x = 40 \sqrt{3} \times 2$

$:\implies\sf \sqrt{3} x = 80 \sqrt{3}$

$:\implies\sf x = \dfrac{80 \cancel{\sqrt{3}}}{ \cancel{ \sqrt{3}}}$

$:\implies{\underline{\boxed{\sf{\pink{x = 80\;m}}}}}\;\bigstar$

$\therefore\; {\underline{\sf{Hence,\;the\; length\;of\;string\;is\; \bf{100\;m}}}}$

Answer 2

Given:-

Height = 40√3m

Angle of elevation = 60°

To find:-

Length of the string.

Answer:-

Let the length of the string be 'x'.

In triangle ABC,

$\tt{sin \: 60 = \frac{Perpendiular}{Height} = \frac{AC}{AB}}$

$\tt {\frac{ \sqrt{3} }{2} = \frac{40 \sqrt{3} }{x}}$

$\tt {\sqrt{3}x = 40 \sqrt{3} \times 2}$ [by cross - multipying on both sides]

$\tt {x = 40 \times 2}$ [since, √3 on both sides get cancelled]

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large\boxed{x = 80m}$

Therefore, Length of the string is 80m.