Question 14 Marks
Kite Festival
Kite festival is celebrated in many countries at different times of the year. In India, every year $14^{\text {th }}$ January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites. The picture given below, three kites flying together.
In Fig. $5,$ the angles of elevation of two kites $($Points $A$ and $B)$ from the hands of a man $($Point $C)$ are found to be $30^{\circ}$ and $60^{\circ}$ respectively. Taking $AD =50 m$ and $BE =60 m,$ find
$(1)$ the lengths of strings used $($take them straight$)$ for kites $A$ and $B$ as shown in the figure.
$(2)$ the distance $'d\ '$ between these two kites
Kite festival is celebrated in many countries at different times of the year. In India, every year $14^{\text {th }}$ January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites. The picture given below, three kites flying together.
In Fig. $5,$ the angles of elevation of two kites $($Points $A$ and $B)$ from the hands of a man $($Point $C)$ are found to be $30^{\circ}$ and $60^{\circ}$ respectively. Taking $AD =50 m$ and $BE =60 m,$ find
$(1)$ the lengths of strings used $($take them straight$)$ for kites $A$ and $B$ as shown in the figure.
$(2)$ the distance $'d\ '$ between these two kites
Answer
$1$. In $\triangle A D C, \angle D=90^{\circ}$
$\sin 30^{\circ}=\frac{A D}{A C}$
$\therefore \frac{1}{2}=\frac{50}{A C}$
or, $ A C=100 m$
In $\triangle B E C, \angle E=90^{\circ}$
$\sin 60^{\circ}=\frac{B E}{B C}$
$\therefore \frac{\sqrt{3}}{2}=\frac{60}{B C}$
or, $ B C=\frac{120}{\sqrt{3}}=40 \sqrt{3} m$
Hence, the length of strings used for kites $A$ and $B$ are $100 m$ and $40 \sqrt{3} m,$ respectively.
$2$. Here, $\angle D C A+\angle A C B+\angle B C E=180^{\circ}$
$($Angles in straight line$)$
$\therefore 30^{\circ}+\angle A C B+60^{\circ}=180^{\circ}$
or, $\angle A C B=180^{\circ}-90^{\circ}=90^{\circ}$
Now, in right $\triangle A C B,$
$A B^2=A C^2+B C^2$
$\Rightarrow d^2$
$=(100)^2+(40 \sqrt{3})^2$
$[$from eq $(i)$ and eq $(ii)]$
$\Rightarrow d^2=10,000+4,800$
$\Rightarrow d^2=14800$
$\Rightarrow d=20 \sqrt{37} \ cm$
Hence, distance between two kites $A$ and $B$ is $20 \sqrt{37} \ cm$.
View full question & answer→$1$. In $\triangle A D C, \angle D=90^{\circ}$
$\sin 30^{\circ}=\frac{A D}{A C}$
$\therefore \frac{1}{2}=\frac{50}{A C}$
or, $ A C=100 m$
In $\triangle B E C, \angle E=90^{\circ}$
$\sin 60^{\circ}=\frac{B E}{B C}$
$\therefore \frac{\sqrt{3}}{2}=\frac{60}{B C}$
or, $ B C=\frac{120}{\sqrt{3}}=40 \sqrt{3} m$
Hence, the length of strings used for kites $A$ and $B$ are $100 m$ and $40 \sqrt{3} m,$ respectively.
$2$. Here, $\angle D C A+\angle A C B+\angle B C E=180^{\circ}$
$($Angles in straight line$)$
$\therefore 30^{\circ}+\angle A C B+60^{\circ}=180^{\circ}$
or, $\angle A C B=180^{\circ}-90^{\circ}=90^{\circ}$
Now, in right $\triangle A C B,$
$A B^2=A C^2+B C^2$
$\Rightarrow d^2$
$=(100)^2+(40 \sqrt{3})^2$
$[$from eq $(i)$ and eq $(ii)]$
$\Rightarrow d^2=10,000+4,800$
$\Rightarrow d^2=14800$
$\Rightarrow d=20 \sqrt{37} \ cm$
Hence, distance between two kites $A$ and $B$ is $20 \sqrt{37} \ cm$.
