Question

A boy is standing on the ground and flying a kite with a string of length 150m at an angle of elevation of 30 degree. Another boy is standing on the roof of a 25m high building and is flying his kite at an elevation of 45 degree. Both the boys are on the opposite sides of both the kites then find the length of the string( in meters) that the second boy must have so that the two kites meet.
2 points

Answer 1

$\huge \bf \pink{ \underline{ \underline{Answer}}}$

• Let the length of second string be x m.

• In ∆ABC

$\bold{ sin \: 30 \degree \: = \frac{AC}{ AB} }$

$\bold{ \frac{1}{2} = \frac{AC }{100} \implies \: AC = 50m}$

• In ∆AEF

$\bold{sin \: 40 \degree = \frac{AF }{AE} }$

$\bold{ = \frac{1}{ \sqrt{2} } = \frac{AC - FC}{x} }$

$\bold{ \frac{1}{ \sqrt{2} } = \frac{50 - 10}{x} } \: \: \\ [ \because \: AC = 50m,FC=ED=10m]$

$\bold{ \frac{1}{ \sqrt{2} } = \frac{40}{x} }$

$\bold{x = 40 \sqrt{2} m}$

$\therefore \: The \: length \: of \: string \: that \\ the \: second \: boy \: must \: have \: so \: that \\ the \: two \: kites \: meet=40 \sqrt{2m}$

Answer 2

$\huge \bf \purple{ \underline{ \underline{Answer}}}$

Let the length of second string be x m.

In ∆ABC

$\sf{ sin \: 30 \degree \: = \frac{AC}{ AB} }$

$\sf{ \frac{1}{2} = \frac{AC }{100} \implies \: AC = 50m}$

In ∆AEF

$\sf{sin \: 40 \degree = \frac{AF }{AE} }$

${ = \frac{1}{ \sqrt{2} } = \frac{AC - FC}{x} }$

$\sf{ \frac{1}{ \sqrt{2} } = \frac{50 - 10}{x} } \: \: \\ \sf[ \because \: AC = 50m,FC=ED=10m]$

$\sf{ \frac{1}{ \sqrt{2} } = \frac{40}{x} }$

$\sf{x = 40 \sqrt{2} m}$

$\sf\therefore \: The \: length \: of \: string \: that \\ \sf \: the \: second \: boy \: must \: have \: so \: that \\ \sf the \: two \: kites \: meet=40 \sqrt{2m}$