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

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 ?

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

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

thanks alot sir

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

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

AB coin=10cm  ∠ADB=(1/2)°  AC=?  tan (1/4)°=((AB/2)/(AC))  ((1/4))°=(1/(4×180))×π radians  tan x≈x for very small x  AC=((AB)/(2×(π/(4×180))))=((10×360)/π)=((3600×7)/(22))cm  =1145cm=11.45m

$$\mathrm{AB}\:\mathrm{coin}=\mathrm{10cm} \\ $$$$\angle{A}\mathrm{DB}=\left(\mathrm{1}/\mathrm{2}\right)° \\ $$$${A}\mathrm{C}=? \\ $$$$\mathrm{tan}\:\left(\mathrm{1}/\mathrm{4}\right)°=\frac{\mathrm{AB}/\mathrm{2}}{\mathrm{AC}} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{4}}\right)°=\frac{\mathrm{1}}{\mathrm{4}×\mathrm{180}}×\pi\:\mathrm{radians} \\ $$$$\mathrm{tan}\:{x}\approx{x}\:\mathrm{for}\:\mathrm{very}\:\mathrm{small}\:{x} \\ $$$$\mathrm{AC}=\frac{\mathrm{AB}}{\mathrm{2}×\frac{\pi}{\mathrm{4}×\mathrm{180}}}=\frac{\mathrm{10}×\mathrm{360}}{\pi}=\frac{\mathrm{3600}×\mathrm{7}}{\mathrm{22}}{cm} \\ $$$$=\mathrm{1145cm}=\mathrm{11}.\mathrm{45m} \\ $$

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

Answered by mrW1 last updated on 29/Nov/17

≈((10)/(0.5))×((180)/π)≈20×60=1200 cm=12m

$$\approx\frac{\mathrm{10}}{\mathrm{0}.\mathrm{5}}×\frac{\mathrm{180}}{\pi}\approx\mathrm{20}×\mathrm{60}=\mathrm{1200}\:{cm}=\mathrm{12}{m} \\ $$

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