The moon's distance from the earth is $360000 k m s$ and its diameter subtends an angle of $31^{'}$ at the eye of the observer. Find the diameter of the moon.

3246.33 k m s

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3747.62 k m s

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3400.52 k m s

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3244.72 k m s

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Solution

The correct option is A $3246.33 k m s$
$A B = 2 d$
$P Q = 36000$
$1^{o} = 60^{^{'}}$
$\Rightarrow \frac{π}{180} \Rightarrow 60^{^{'}}$
$\Rightarrow 1^{^{'}} \to \frac{π}{180 \times 60}$
$∴ 31^{^{'}} \to \frac{31 π}{180 \times 60} r a d i u s$
$△ A P Q ≅ △ P Q B$
$∴ L P Q A = L P Q B = \frac{31 π}{180 \times 120}$ radius
tan $(L P Q A) = \frac{A P}{P Q} = \frac{d}{P Q}$
$\to 4.5088 = \frac{d}{36 \times 10^{4}}$
$\to d = 1623.168$
$∴$ diameter $2 d = 3246.33 k m$

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The moon's distance from the earth is 360000 kms and its diameter subtends an angle of 31′ at the eye of the observer. Find the diameter of the moon.

The correct option is A 3246.33 kmsAB=2d PQ=360001o=60′⇒π180⇒60′ ⇒1′→π180×60 ∴31′→31π180×60radius △APQ≅△PQB∴LPQA=LPQB=31π180×120 radiustan (LPQA)=APPQ=dPQ→4.5088=d36×104→d=1623.168∴ diameter2d=3246.33km

The moon's distance from the earth is $360000 k m s$ and its diameter subtends an angle of $31^{'}$ at the eye of the observer. Find the diameter of the moon.