Question

a satellite orbiting the earth passes directly overhead at observation stations in bengaluru and sriharikkotta. at an instant when satellite is between these two stations its angle of elevation is simultaneously observed to be complementary angles. if distance between bengabengaluru and sriharikkotta is 350km.find •value of cot theta + tan theta in terms of altitude of satellite from earth. • if angle of elevations are equal complementary angles. find altitude of satellite​

Answer 1

Given :  satellite  between bengaluru and sriharikkotta   , angle of elevation is simultaneously observed to be complementary angles and distance between stations is 350 km

To find : Tanθ + Cotθ  in terms of altitude

Solution:

Let say  altitude of satellite from earth = H

& point of altitude as at distance of x from station making angle θ

Tanθ   =  h/x    & Cotθ = x/h

Tan(90 - θ)  = h/(350 - x)   & Cot(90-θ) = (350 - x)/h

=> Cotθ = h/(350 - x)    & Tanθ = (350 - x)/h

=> h/x  = (350 - x)/h

=> h² = 350x - x²

=> x² + h² = 350x

or  as these two angle are complementary  hence angle made by both stations  at satellite = 90°

Hence  x² + h²   + (350-x)²  + h²   = 350²

=> x² + h² + 350² + x²  - 700x  + h² = 350²

=> 2x² +  2h²  = 700x

=> x² + h² = 350x

Tanθ + Cotθ

= h/x  + x/h

= (h² + x²)/xh  

= 350x/xh

= 350/h

Tanθ + Cotθ  = 350/h

angle of elevations are equal complementary angles

then  θ  = 1  & Cotθ  = 1

=> 1 + 1  = 350/h

=> h = 175

Then altitude of satellite​ = 175 km

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