If projections of three points `A, B, C` on a given plane are `A′,B′,C′` ; then `ΔA′B′C′=cosθ(ΔABC)` , w

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Q 1277134086.     If projections of three points `A, B, C` on a given plane are `A′,B′,C′` ; then `ΔA′B′C′=cosθ(ΔABC)` , where `θ` is the angle between the planes `ABC` and `′B′C′` . In general, if `A_0` is the area of any plane curve and `A` is the area of its projection on any plane, then `A=cosθA_0`

Suppose `AB` is a diameter of a circle and `P` is a plane through `AB` making an angle `θ` with the plane of the circle. If diameter of the circle is `2a`, then the eccentricity of the curve of projection of the circle on `P` is
A

`sin θ`

B

`(2asinθ)/(1+a)`

C

`(2acosθ)/(1+a)`

D

`cos θ`

HINT


Solution
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

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