Question

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 10m high building and is flying a kite at an elevation of 45 °. Both the boys are on the opposite sides of the kites. Find the length of the string that the second boy must have so that the two kites meet.

Answer 1

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

Suppose, length of the second kite= L meters

Height of 2nd kite=  L x Sin 45 + 10 m 
L sin 45 + 10 = 50 m
L = 40√2 m

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 10m 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= 10m
• 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° = $\frac{BE+EF}{AF}$

→ $\frac{1}{2}=\frac{10+EF}{100}$

→ 2(10 + EF) = 100

→ 20 + 2EF = 100

→ 2EF = 100 -20

→ 2EF = 80

→ EF = $\cancel{\frac{80}{2} }$

→ EF = 40

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

→ sin45° = $\frac{EF}{DF}$

→ $\frac{1}{\sqrt{2} } =\frac{40}{DF}$

→ DF = 40× √2

→ DF = 40√2 m

Thus,

The length of the string that the second boy must have so that the two kites meet is 40√2 m.