Question

Find the magnitude of the resultant of the concurrent forces of 8 N, 12 N, 15N and 20 N making angles of 30°, 70°, 120°and 155° respectively with a fixed-line.


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Answer 1

Answer:

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Answer 2

Answer:

The resultant of two vectors:

P and Q is given by

r = \sqrt{p2 + q2 + 2pq \cos(θ) }

r=p2+q2+2pqcos(θ)

where, P and Q are magnitudes of two

vectors and θ is the angle between two

vectors.

Here θ=? P=8N Q=15N R=17N

Therefore,

17 = \sqrt{ {8}^{2} + {15}^{2} + 2 \times 8

\times 15 \ \cos(θ) }17=82+152+2×8×15 cos(θ)

289= 64 + 225 + 240 cosθ

289= 289 + 240 cosθ

289 - 289 = 240 cosθ

cosθ = 0

Step-by-step explanation:

PLZZ MARK AS BRAINLIEST(and thanks for the answer)