Question

an observer spots a balloon moving with the wind in a horizontal line at a height of 105 m from the ground. the angle of elevation of the balloon after spmetime reduces from 60 degree to 30 degree.find the distance travelled by the balloon during the interval​

Answer 1

Answer:

$x = 70\sqrt{3} m$  

Step-by-step explanation:

Let the distance travelled by the balloon be x metres

Let the initial distance between the balloon and observer be y metres

Let the final distance between the balloon and observer be z metres

$tan60=\sqrt{3} = \frac{105}{y}$

$y=\frac{105}{\sqrt{3} } m$

$tan 30 = \frac{1}{\sqrt{3} } = \frac{105}{z}$

$z=105\sqrt{3} m$

$x=z-y=105[\sqrt{3} -\frac{1 }{\sqrt{3} } ]$  

$x=105[\frac{2 }{\sqrt{3} } (\frac{\sqrt{3} }{\sqrt{3} } )]$

$x = 70\sqrt{3} m$

HENCE SOLVED

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