Question

A boy is standing on the ground and flying kite with 100m of string at an elevation of 30 degree . Another boy is standing on the roof of a 20m 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

Answer:

The height of the first kite above ground = 100 m * Sin 30 = 50 meters

Let the length of the second kite be : L meters

Height of 2nd kite:  L Sin 45 + 10 meters

So L sin 45 + 10 = 50 m

 => L = 40√2 meters

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Answer 2

$\bf{\Huge{\underline{\boxed{\sf{\green{ANSWER\::}}}}}}$

Given:

A boy standing on the ground and flying a kite with 100m of string at an elevation of 30°. Another boy is standing on the roof of a 20m high building & is flying kite at an elevation of 45°. Both the boys are on opposite sides of both the kites.

$\bf{\Large{\underline{\rm{\red{To\:find\::}}}}}$

The length of the string that the second boy must have so that the two kites meet.

$\bold{\Large{\underline{\sf{\pink{Explanation\::}}}}}$

In figure,

• Let A be the boy & AF is the position of kite.

• Let BF be vertical height of the kite= 20m
• Length of the string,AF = 100m
• Angle of elevation of the kite,∠BAF = 30°

In right angled ΔABF,

$\bold{\sin\theta=\frac{Perpendicular}{Hypotenuse} }}$

→ sin30° = $\frac{BF}{AF}$

→ sin30° = $\frac{BE+EF}{AF}$

→ $\frac{1}{2}=\frac{20+EF}{100}$

→ 2(20 + EF) = 100

→ 40 + 2EF = 100

→ 2EF = 100 -40

→ 2EF = 60

→ EF = $\cancel{\frac{60}{2} }$

→ EF = 30

Let D be the position of second boy& DF be the length of second kite. given ∠EDF = 45°

→ sin45° = $\frac{EF}{DF}$

→ $\frac{1}{\sqrt{2} } =\frac{30}{DF}$

→ DF = 30× √2

→ DF = 30√2 m

Thus,

The length of the string that the second boy must have so that the two kites meet is 30√2 m.