Question

1. A kite is flying at a height of 75m
from the level ground, attached to a
string inclined at 60° to the
horizontal. Find the length of the
string?​

Answer 1

${\bold{\huge{\blue{\underline{\red{Gi}\pink{ve}\green{n}\purple{:-}}}}}}$

• AB = 75 m

• ∠ ACB = 60°

${\bold{\huge{\pink{\underline{\pink{F}\purple{in}\blue{d}\red{:-}}}}}}$

• Length of the String (AC) = ?

${\bold{\huge{\pink{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}}$

In ∆ ABC

$\bf\dfrac{AB}{AC}=\dfrac{P}{H}=\:Sin\:60\degree$

$\bf\implies\dfrac{75}{AC}=\dfrac{\sqrt{3}}{2}$

$\bf\implies\:AC\:\sqrt{3}=150$

$\bf\implies\:AC=\dfrac{150}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}$

$\bf\implies\:AC=\dfrac{150\sqrt{3}}{3}$

$\bf\implies\:AC=50\sqrt{3}\:m$

$\bf\implies\:AC=50\times\:1.732$

$\bf\implies\:AC=86.6\:m$

Hence, the length of the String is 86.6 m

Answer 2

Given :

• Height of kite = 75 m

• Angle of $\theta$ = 60°

To Find :

• Length of string

Solution :

• Let's the height of kite is perpendicular height of triangle and string be the hypothenuse of triangle with theta 60°

By Using

$\boxed{ \boxed{ \sf \large \sin \theta = \dfrac{p}{h}}}$

Substitute value in formula

$\sf \implies \sin60 = \frac{75}{h} \\ \\ \sf \implies \frac{ \sqrt{3} }{2} = \frac{75}{h} \\ \\ \sf \implies \sqrt{3} h = 150 \\ \\ \sf \implies h = \frac{150}{ \sqrt{3} }$

By rationalisation of denominator

$\sf \implies h = \frac{150}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ \sf \implies h = \frac{150 \sqrt{3} }{3} \\ \\\sf \implies h = 50 \sqrt{3} \\ \\ \implies \large \boxed{ \boxed{ \sf h =86.6\:m} }$