Find the diameter of the sun in km supposing that it subtends an angle of 32 at the eye of an observer. Given that the distance of the sun is 91 x 106 km.

Explanation

Let `P Q` be the diameter of the sun and `E` be the eye of the observer.


The distance between the Sun and Earth is large, so we will take `PQ` as arc `PQ`.


Given radius `=91 times 10^{6} km`


`Rightarrow theta=32^{prime}=(frac{32}{60})^{circ}=(frac{32}{60} times frac{pi}{180}) rad`


We know that `theta=frac{text { arc }}{text { radius }}`



`Rightarrow frac{32}{60} times frac{pi}{180}=frac{d}{91 times 10^{6}}`


`therefore {d}=frac{32 times 91 times 10^{6} times 22}{60 times 180 times 7}


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