In a triangle- if r1 - r3 - k cos2B-2-then k is equal to-

Determine the value of k Calculating r_1 + r_3 :: r_1 = 4R sin(A/2) cos(B/2)cos(C/2)0ex0exr_3 = 4R cos(A/2) cos(B/2) sin(C/2)0ex0exaddingr_1andr_3, weget0ex0ex ...

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Byju's Answer

Standard XII

Mathematics

Quotient Rule of Differentiation

In a triangle...

Question

In a triangle, if $r_{1} + r_{3} = k \cos^{2} (\frac{B}{2}),$ then $k$ is equal to:

$R$

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$2 R$

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$3 R$

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$4 R$

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Solution

The correct option is D
$4 R$

Determine the value of $k$
Calculating $r_{1} + r_{3} :$
: $r_{1} = 4 R \sin (\frac{A}{2}) \cos (\frac{B}{2}) \cos (\frac{C}{2}) r_{3} = 4 R \cos (\frac{A}{2}) \cos (\frac{B}{2}) \sin (\frac{C}{2}) adding r_{1} and r_{3}, we get r_{1} + r_{3} = 4 R \sin (\frac{A}{2}) \cos (\frac{B}{2}) \cos (\frac{C}{2}) + 4 R \cos (\frac{A}{2}) \cos (\frac{B}{2}) \sin (\frac{C}{2}) = 4 R \cos (\frac{B}{2}) (\sin (\frac{A}{2}) \cos (\frac{C}{2}) + \cos (\frac{A}{2}) \sin (\frac{C}{2})) = 4 R \cos (\frac{B}{2}) \sin (\frac{A + C}{2}) = 4 R \cos (\frac{B}{2}) \sin (\frac{180 ° - B}{2}) r_{1} + r_{3} = 4 R \cos^{2} (\frac{B}{2}) Comparing with r_{1} + r_{3} = k c o s^{2} (\frac{B}{2}) [Given] ∴ k = 4 R$
Hence, Option (D) is correct.

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In a triangle, if r1+r3=kcos2B2,then k is equal to:

In a triangle, if $r_{1} + r_{3} = k \cos^{2} (\frac{B}{2}),$ then $k$ is equal to: