Two forces are such that the sum of their magnitude is 18N and their resultant which has magnitude 12N, is perpendicular to the smaller force. Then the magnitudes of the forces are

12N, 6N

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13N, 5N

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10N, 8N

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16N, 2N

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Solution

The correct option is B
13N, 5N

$A + B = 18$ ------(1)
$12^{2} = A^{2} + B^{2} + 2 A B c o s θ$
$t a n α = \frac{B s i n θ}{A + B c o s θ} \Rightarrow A + B c o s θ = 0$
or $c o s θ = - A / B$
$144 = A^{2} + B^{2} - 2 A^{2} o r B^{2} - A^{2} = 144$ --------(2)
Solving(1)&(2)B=13N; A=5N

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Two forces are such that the sum of their magnitude is 18N and their resultant which has magnitude 12N, is perpendicular to the smaller force. Then the magnitudes of the forces are

The correct option is B 13N, 5N A+B=18 ------(1) 122=A2+B2+2AB cosθ tanα=BsinθA+Bcosθ⇒A+Bcosθ=0 or cosθ=−A/B 144=A2+B2−2A2orB2−A2=144 --------(2) Solving(1)&(2)B=13N; A=5N

The correct option is B
13N, 5N

$A + B = 18$ ------(1)
$12^{2} = A^{2} + B^{2} + 2 A B c o s θ$
$t a n α = \frac{B s i n θ}{A + B c o s θ} \Rightarrow A + B c o s θ = 0$
or $c o s θ = - A / B$
$144 = A^{2} + B^{2} - 2 A^{2} o r B^{2} - A^{2} = 144$ --------(2)
Solving(1)&(2)B=13N; A=5N