RRB NTPC Mains or Stage 2 Model Paper 7 with solutions

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Question : 40 of 120

Marks: +1, -0

If

P sin θ= √3 and P cos θ= 1

, then the value of P is

$2 /√3$
$-1 /√3$
2
$1/ 2$

Solution:

(c)

P sin θ= 3 ...(i)

P cos θ= 1 ...(ii)

From Eqs. (i) and (ii),

{Psin θ}/{Pcosθ} =√3

⇒tan θ= √3

tan θ= tan 60°

∴θ= 60°

From Eq. (i),

P sin θ= √3 ⇒P sin 60° = √3

⇒p√3/2=√3

∴p=2

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