Question

From a balloon vertically above a straight road, the angles of depression of two cars at an instant are found to be 45° and 60°.lf the cars are 100m apart, find the height of the balloon.......Please don't copy and answer your own.... ​

Answer 1

Answer:

Explanation:

Solution:-

Let the two angle be x = 45° and y = 60° respectively.

And the distance between the two cars is d = 100 m

Height of balloon is AB  = h meter (See the figure)

Let BC = x meter

In triangle ABO,

tan 45° = AB/BO  ⇒ 1 = h/(x+100) ⇒ h = x + 100 ........(1)

In triangle ABC,

tan 60° = AB/BC  

⇒ √3 = h/x

⇒ h = x√3 ...... (2)

As h = x + 100

⇒ x + 100 = x√3

⇒ 100 = √3 x - x

⇒ 100 = x(√3 - 1)

⇒ x = 100/(√3 - 1)

⇒ x = 100 (√3 + 1)/(√3 - 1) (√3 + 1)/(√3 + 1)    {Multiplied by √3 +1 with both numerator and denominator}

⇒ x = 100(1.73 + 1)/2  

⇒ x = 100(2.73/2)

⇒ x = 273/2

⇒ x = 136.5 m

So, BC = 136.5 meter

So, h = x + 100

h = 136.5 + 100

h = 236.5 meter

Answer 2

Answer:

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