Math, asked by Shivaniainapur1385, 10 months ago

Satellite orbiting Earth passes directly overhead at observation stations in Bangalore and Sriharikota at an instant when satellite is in between these two stations its angle of elevation is simultaneously observed by to be complementary angles if the distance between Bangalore and Sriharikota is 350 kilometre find value of cot theta + tan theta in terms of altitude of satellite from Earth if angle of elevation side equal complementary angles find altitude of satellites and the condition is that angle angle angle of elevation from Sriharikota is 90 - theta

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Answers

Answered by amitnrw
2

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 is at distance of x from station (bangalore) making angle θ  => Distance of that point from Sriharikota  = 350 - x

Tanθ   =  h/x    => cotθ = x/h

Tan(90 - θ)  = h/(350 - x)  

=> Cotθ = h/(350 - x)  

=> 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 ( applying Pythagoras theorm )

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

=> 2x² +  2h²  = 700x

=> x² + h² = 350x

Tanθ + Cotθ

= h/x  + x/h

= (h² + x²)/xh  

using x² + h² = 350x

= 350x/xh

= 350/h

Tanθ + Cotθ  = 350/h

angle of elevations are equal complementary angles

=> θ = 90° - θ

=> 2θ = 90°

=> θ = 45°

tan 45°  = 1  & Cot45°  = 1

Tanθ + Cotθ  =  1 + 1 = 2

=>2  = 350/h

=> h = 175

Then altitude of satellite​ = 175 km

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