11. The sun angular diameter is …
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Arun Soni 8 years, 5 months ago
{tex}\eqalign{ & Given{\text{ D = 1}}{\text{.496}} \times {\text{1}}{{\text{0}}^{11}}m \cr & \theta = {19^ \circ }{20^{''}} = {(\frac{{1920}}{{60 \times 60}})^{^ \circ }} = {0.53^{^ \circ }} \cr & {\text{converting it to radian}} \cr & {\text{0}}{\text{.53}} \times \frac{\pi }{{180}} = 0.53 \times \frac{{3.142}}{{180}} = 0.0093{\text{ rad}} \cr & As{\text{ }}\theta {\text{ is too small, we can assume triangle as right angle triangle}} \cr & {\text{we know d = D tan}}\theta \cr & {\text{As }}\theta {\text{ is small we can take tan}}\theta \approx \theta \cr & \Rightarrow {\text{d = D}}\theta \cr & \therefore d = 1.496 \times {10^{11}} \times 0.0093 = 1.39 \times {10^9}{\text{ m}} \cr} {/tex}
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