Question

The angular diameter is measured to be 1920''. The distance D of the sun from the Earth is 1.496× 10 to the power 11 m. what is the diameter of the sun??​

Answer 1

Given:

D = 1.496 × 1011 m

theta = 1920 seconds

 

Solution:

 

The distance D from the Earth to the Sun is very large.

This makes the diameter of the Sun 'd' almost like a straight line.

Also, the value of theta also gets reduced due to the the very large distance of 'D'.

Such an orientation makes it a right angled triangle.

 

Using the definition of trigonometric ratios we get,

 

tan (theta) = d / D ... (1)

As theta is negligible, tan (theta) is also redundant.

 

Thus  

tan (theta) = theta ...(2)

Applying (2) in (1)

 

(theta) = d / D ...(3)

 

converting theta from seconds to degrees,

theta in degrees = 1920 seconds / (60 x 60) = 0.53

 

converting theta in degrees to radians for calculation,

theta in radians = 0.53 x π/180 = 0.0093  

 

Using (3)

d = D x theta in radians

d= 1.496 × 1011 m x 0.0093 radians

d= 1.39 x 109 m

Answer 2

Answer:

Diameter of the sun = 1.39 x 109 m

Explanation:

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The distance D from the Earth to the Sun is very large.

This makes the diameter of the Sun 'd' almost like a straight line.

Also, the value of theta also gets reduced due to the the very large distance of 'D'.

Such an orientation makes it a right angled triangle.

Using the definition of trigonometric ratios we get,

tan (theta) = d / D ... (1)

As theta is negligible, tan (theta) is also redundant.

Thus  

tan (theta) = theta ...(2)

Applying (2) in (1)

(theta) = d / D ...(3)

converting theta from seconds to degrees,

theta in degrees = 1920 seconds / (60 x 60) = 0.53

converting theta in degrees to radians for calculation,

theta in radians = 0.53 x π/180 = 0.0093  

Using (3)

d = D x theta in radians

d= 1.496 × 1011 m x 0.0093 radians

d= 1.39 x 109 m