Question

A kite is flying at a height of 30m above the ground.the string attacked to the kite as tempprary tied to a point om tje groumd is 45 degree.Find the lengh of the string.

Answer 1

✬ Length = 42.42 m ✬

Step-by-step explanation:

Given:

• A kite is flying at a height of 30 m.
• Inclination of string with the ground is 45°.

To Find:

• What is the length of string ?

Solution: Let AB be height 30 m above ground , AC be the string of kite and ∠ACB be the inclination of string towards ground.

➽ Also,AB is perpendicular to BC, ∠ABC = 90°

Now, in right triangle ABC

$\implies{\rm }$ sin θ = Perpendicular/Hypotenuse

$\implies{\rm }$ sin C = AB/AC

$\implies{\rm }$ sin 45° = 30/AC

$\implies{\rm }$ 1/√2 = 30/AC

$\implies{\rm }$ AC = 30√2

$\implies{\rm }$ AC = 30(1.414)

$\implies{\rm }$ AC = 42.42 m

Hence, the length of string of kite is 42.42 m.

Answer 2

Step-by-step explanation:

$\sf \red {\underline{\underline{Given:-}}}$

• Height of the kite from the ground = 30m

• The kite is inclined at an angle of 45° from the ground

$\sf \blue {\underline{\underline{To\:Find:-}}}$

• The length of the string

$\sf \green {\underline{\underline{Solution:-}}}$

$\sf \pink{Let \: the \: length \: \: of \: \: the \: string \: be \: (l)}$

P is the position of the kite,

Q is the ground point .

PQ = 30m

$\sf{Given \: \angle POQ = 45 \degree}$

$\rightarrow\sf \sin \theta = \dfrac{Opposite \: \: side}{</p><p>Hypotenuse}$

$\rightarrow\sf \sin 45\degree= \dfrac{PQ}{OP}$

$\rightarrow\sf \sin 45 \degree= \dfrac{30}{l}$

$\rightarrow \sf As \: we \: know \: that \: \sin 45 \degree = \dfrac{1}{ \sqrt{2} }$

$\rightarrow \sf \dfrac{1}{ \sqrt{2} } = \dfrac{30}{l}$

By cross multiplication:-

$\rightarrow \sf l= 30 \times \sqrt{2}$

$\rightarrow \sf l= 30 \times (1.414)$

$\sf \rightarrow \:l = 42.42m$

Therefore:-

$\boxed{ \sf { \purple{ Length \: \: of \: \: the \: \: string= 42.42m}}}$