Question

A boy is standing on the ground and flying kite with 120m of string at an elevation of 30 degree . another boy is standing on the roof of a 14m high high building and is flying his kite at an elevation of 45 degree. both these boys are on opposite sides of both the kites. find the length of the string that the second boy must have so that the two kites meet

Answer 1

Step-by-step explanation:

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Answer 2

The length of the string that the second boy must have $46\sqrt 2$ m so that the two kite meet.

Step-by-step explanation:

AB=120 m

EFCD is a rectangle therefore,

DE=FC=14 m

AC=AF+FC=AF+14

Let AF=x

AC=x+14

In triangle ABC

$\frac{Perpendicular\;side}{Hypotenuse}=Sin\theta$

$\frac{AC}{AB}=sin 30^{\circ}$

$\frac{x+14}{120}=\frac{1}{2}$

$sin 30=\frac{1}{2}$

$x+14=60$

$x=60-14=46$

AF=46 m

In triangle AFE

$\frac{AF}{AE}=sin 45$

$\frac{46}{AE}=\frac{1}{\sqrt 2}$

$sin 45=\frac{1}{\sqrt 2}$

$AE=46\sqrt 2m$

Hence, the length of the string that the second boy must have $46\sqrt 2$ m so that the two kite meet.

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https://brainly.in/question/213770 :Answered by Kvnmurty