Question

A string of length 50√3 m attached to a kite is temporarily tied to a point on the ground at an inclination ϴ. Find the angle of inclination of the string if the kite is flying at the height of 75 m above the ground, assuming that there is no slack in the string.
(Class 10 Maths Sample Question Paper)

Answer 1

FIGURE IS IN THE ATTACHMENT
Given :
Height of kite (AB) = 75 cm
Length of string (AC) = 50√3 cm

In ∆ ABC ,
sin ϴ = AB / AC
[ sin ϴ = perpendicular/ hypotenuse]

sin ϴ = 75 / 50√3
sin ϴ = 3 / 2√3
sin ϴ = (3 × √3) / (2√3 × √3)

[ By rationalising the denominator ]
sin ϴ = 3√3 / (2 ×3) = 3√3 / 6 = √3 /2
sin ϴ = √3 / 2
sin ϴ = sin 60 °
ϴ = 60 °

Hence, the angle of inclination on the string is 60 ° .

HOPE THIS WILL HELP YOU....

Answer 2

HELLO DEAR,


LET THETA=A

$\sin(a) = \frac{kp}{ks} \\ = > \sin(a) = \frac{75}{50 \sqrt{3} } \\ = > \sin(a) = \frac{75 }{50 \times \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ = > \sin(a) = \frac{3}{2 \times \sqrt{3} } \ \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ = > \sin(a) = \frac{ \sqrt{3} }{2} \\ = > \sin(a) = \sin(60) \\ = > a = 60$


I HOPE ITS HELP YOU DEAR,
THANKS