Question

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 metres from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is ${60}^{\circ }$. After some time, the angle of elevation reduces to ${30}^{\circ }$. Find the distance traveled by the balloon during the interval.
[4 MARKS]

Answer 1

Steps: 3 Marks
Answer: 1 Mark

Let AB be the position of the girl and AX be the horizontal ground. Let C and D be the two positions of the balloon.

Draw $CL\perp AX,DM\perp AX$ and $BN\perp DM$, intersecting CL at P.



Then, $\angle CBP={60}^{\circ },\angle DBN={30}^{\circ },AB=PL=NM=1.2m$ and $CL=DM=88.2m$

$\therefore$ CP = 88.2 m - 1.2 m = 87 m

From right $\Delta BPC$, we have

$\frac{BP}{CP}=cot{60}^{\circ }=\frac{1}{\sqrt{3}}\Rightarrow \frac{BP}{87m}=\frac{1}{\sqrt{3}}\Rightarrow BP=\frac{87m}{\sqrt{3}}$

$\Rightarrow BP=\frac{87m}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=29\sqrt{3}m$

From right $\Delta BND$, we have

$\frac{BN}{DN}=cot{30}^{\circ }=\sqrt{3}\Rightarrow \frac{BP+PN}{CP}=\sqrt{3}$

$\Rightarrow \frac{29\sqrt{3}+CD}{87m}=\sqrt{3}[\because PN=CDandDN=CP]$

$\Rightarrow 29\sqrt{3}m+CD=87\sqrt{3}m$

$\Rightarrow CD=87\sqrt{3}m-29\sqrt{3}m=58\sqrt{3}m$

Hence, the required distance traveled by the balloon is $58\sqrt{3}m$