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
Answers
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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