Assuming-that-the-moon-s-diameter-subtends-and-angle-1-2-at-the-eye-of-an-observer-find-how-far-from-the-eye-of-a-coin-of-10-cm-diameter-must-be-held-so-as-just-to-hide-moon-

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Question Number 24948 by adityapratap2585@gmail.com last updated on 29/Nov/17

$$\mathrm{Assuming}\mathrm{that}\mathrm{the}{\mathrm{moon}}^{\prime }\mathrm{s}\mathrm{diameter}$$$$\mathrm{subtends}\mathrm{and}\mathrm{angle}\left(1/2\right)°\mathrm{at}\mathrm{the}\mathrm{eye}$$$$\mathrm{of}\mathrm{an}\mathrm{observer},\mathrm{find}\mathrm{how}\mathrm{far}\mathrm{from}\mathrm{the}$$$$\mathrm{eye}\mathrm{of}\mathrm{a}\mathrm{coin}\mathrm{of}10\mathrm{cm}\mathrm{diameter}\mathrm{must}$$$$\mathrm{be}\mathrm{held}\mathrm{so}\mathrm{as}\mathrm{just}\mathrm{to}\mathrm{hide}\mathrm{moon}?$$

Commented by adityapratap2585@gmail.com last updated on 29/Nov/17

$$\mathrm{thanks}\mathrm{alot}\mathrm{sir}$$

Commented by prakash jain last updated on 29/Nov/17

$$\mathrm{AB}\mathrm{coin}=10\mathrm{cm}$$$$\mathrm{\angle }A\mathrm{DB}=\left(1/2\right)°$$$$A\mathrm{C}=?$$$$\tan \left(1/4\right)°=\frac{\mathrm{AB}/2}{\mathrm{AC}}$$$$\left(\frac{1}{4}\right)°=\frac{1}{4\times 180}\times \pi \mathrm{radians}$$$$\tan x\approx x\mathrm{for}\mathrm{very}\mathrm{small}x$$$$\mathrm{AC}=\frac{\mathrm{AB}}{2\times \frac{\pi }{4\times 180}}=\frac{10\times 360}{\pi }=\frac{3600\times 7}{22}cm$$$$=1145\mathrm{cm}=11.45\mathrm{m}$$

Commented by prakash jain last updated on 29/Nov/17

Answered by mrW1 last updated on 29/Nov/17

$$\approx \frac{10}{0.5}\times \frac{180}{\pi }\approx 20\times 60=1200cm=12m$$

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