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

12 N, 6 N

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13 N, 5 N

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10 N, 8 N

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16 N, 2 N

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Solution

The correct option is B
13 N, 5 N

$A + B = 18 . . . . (i)$
$12 = \sqrt{A^{2} + B^{2} + 2 A B c o s θ} . . . . (i i)$
$t a n α = \frac{B s i n θ}{A + B c o s θ} = t a n 90^{o} \Rightarrow= - \frac{A}{B} . . . . (i i i)$
By solving (i), (ii) and (iii), A=13 N and B=5 N

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

The correct option is B 13 N, 5 N A+B=18 ....(i) 12=√A2+B2+2ABcosθ ....(ii) tan α=B sinθA+B cosθ=tan 90o ⇒=−AB ....(iii) By solving (i), (ii) and (iii), A=13 N and B=5 N

The correct option is B
13 N, 5 N

$A + B = 18 . . . . (i)$
$12 = \sqrt{A^{2} + B^{2} + 2 A B c o s θ} . . . . (i i)$
$t a n α = \frac{B s i n θ}{A + B c o s θ} = t a n 90^{o} \Rightarrow= - \frac{A}{B} . . . . (i i i)$
By solving (i), (ii) and (iii), A=13 N and B=5 N