 Krishna
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Step 1: Read the given question and make a figure using the given information. GIVEN:  1.2 m girl spots a balloon

Girl height AG = BF = 1.2 m ( \because AC // GF)

Height of the balloon EF = 88.2 m

The angle of elevation of the balloon from the eye of the girl

\angle EAB = 60\degree

After some time, angle of elevation

\angle DAC = 30\degree

Horizontal distance traveled by the balloon BC = ?

FROM THE FIGURE:

Height of balloon above the girl height

BE = EF - BF

=  88.2 - 1.2

BE = 87 m

BE = DC = 87 m................(1)

Step 2: Find the possible distances between the points using the trigonometric ratios

From right triangle ABE

\tan\theta=\frac{opposite\ side\ of\ the\ angle}{adjacent\ side\ of\ the\ angle}

\tan 60\degree = \frac{87}{AB}

\sqrt{3} = \frac{87}{AB}              \because \tan 60\degree = \sqrt{3}

AB = \frac{87}{\sqrt{3}}

\tan A = \frac{CD}{AC} = \frac{87}{AC}

\tan 30\degree = \frac{87}{AC}

\frac{1}{\sqrt{3}} = \frac{87}{AC}                             \because \tan 30\degree = \frac{1}{\sqrt{3}}

AC = 87\sqrt{3}

Step 3: Find the horizontal distance traveled by the balloon

From the figure:

AC = AB + BC

BC = AC - AB

BC = 87\sqrt{3} - \frac{87}{\sqrt{3}}

BC = 87 (\sqrt{3} - \frac{1}{\sqrt{3}})

BC = 87 (\frac{\sqrt{3}*\sqrt{3} - 1}{\sqrt{3}})

BC=87\left(\frac{3-1}{\sqrt{3}}\right)

BC = \frac{87*2}{\sqrt{3}}

Multiply \sqrt{3} numerator and denominator

BC = \frac{87 * 2*\sqrt{3}}{\sqrt{3}*\sqrt{3}}

BC = \frac{87*2*\sqrt{3}}{3}

BC = 29*2*\sqrt{3}

BC = 58\sqrt{3}

Hence, distance traveled by the balloon BC = 58\sqrt{3}