Question

Q: Find the length of the perpendicular from the origin to the line joining two points whose coordinates are (cosA,cosB) and (sinA,SinB) .

[NCERT 11TH]

Answer 1

Answer:

P = cos(∅ - Φ)/2

Step-by-step explanation:

equation of line joining the points (cos∅, sin∅) and (cosΦ, sinΦ) is

( y - sin∅) = (sinΦ - sin∅)/(cos∅ - cosΦ) (x -cos∅)

[ use formula , sinC - sinD = 2cos(C+D)/2.sin(C-D)/2 and CosC - cosD = -2sin(C + D)/2.sin(C-D)/2 ]

y - sin∅ = {2cos(∅+Φ)/2.sin(∅-Φ)/2/-2sin(∅+Φ)/2.sin(∅-Φ)/2} (x - cos∅)

y - sin∅ =- {cos(∅+Φ)/2/sin(∅+Φ)/2}(x - cos∅)

{sin(∅+Φ)/2}y - sin∅.sin(∅+Φ)/2 + {cos(∅+Φ)/2}x - cos∅.cos(∅+Φ)/2 = 0

[ cos(A- B ) = cosA.cosB + sinA.sinB use it here ]

{sin(∅+Φ)/2}y + {cos(∅+Φ)/2}x -cos(2∅-∅-Φ)/2 = 0

{sin(∅+Φ)/2}y + {cos(∅+Φ)/2}x -{cos(∅-Φ)/2} =0

it distance from origin (0,0) is

P = | 0 + 0 - cos(∅ - Φ)/2|/√{cos²(∅+Φ)/2 + sin²(∅+Φ)/2 }

[ use formula, P = |ax1 + by1 +c|/√(a² + b²) ]

P = cos(∅ - Φ)/2

hope it will helps you friend

Answer 2

HEYA MATE. ..
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P = cos(∅ - Φ)/2



equation of line joining the points (cos∅, sin∅) and (cosΦ, sinΦ) is

( y - sin∅) = (sinΦ - sin∅)/(cos∅ - cosΦ) (x -cos∅)


[ use formula , sinC - sinD = 2cos(C+D)/2.sin(C-D)/2 and CosC - cosD = -2sin(C + D)/2.sin(C-D)/2 ]


y - sin∅ = {2cos(∅+Φ)/2.sin(∅-Φ)/2/-2sin(∅+Φ)/2.sin(∅-Φ)/2} (x - cos∅)


y - sin∅ =- {cos(∅+Φ)/2/sin(∅+Φ)/2}(x - cos∅)


{sin(∅+Φ)/2}y - sin∅.sin(∅+Φ)/2 + {cos(∅+Φ)/2}x - cos∅.cos(∅+Φ)/2 = 0

[ cos(A- B ) = cosA.cosB + sinA.sinB use it here ]


{sin(∅+Φ)/2}y + {cos(∅+Φ)/2}x -cos(2∅-∅-Φ)/2 = 0


{sin(∅+Φ)/2}y + {cos(∅+Φ)/2}x -{cos(∅-Φ)/2} =0


it distance from origin (0,0) is

P = | 0 + 0 - cos(∅ - Φ)/2|/√{cos²(∅+Φ)/2 + sin²(∅+Φ)/2 }


[ use formula, P = |ax1 + by1 +c|/√(a² + b²) ]


P = cos(∅ - Φ)/2
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HOPE THIS HELPS YOU