    Question

# If F1, F2 makes an angle of 60∘ and 37∘ respectively with F3 as shown in the figure, and magnitude of F3 is 2 N, then the magnitude of F1 and F2 respectively are (Given, −→F1+−→F2=−→F3) A
43(3+4) N and 10(3+4) N
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B
23(3+4) N and 10(3+4) N
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C
43(3+4) N and 5(3+4) N
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D
4(3+4) N and 5(3+4) N
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Solution

## The correct option is A 4√3(√3+4) N and 10(√3+4) NLet the magnitude be |−→F1|=x and |−→F2|=y Also, Given, −→F1+−→F2=−→F3 So, the given vectors can be shown as Now, resolving the vectors along the axes we get |−−→F3x|=xcos60∘+ycos37∘ or, |−−→F3x|=(x2+4y5) Similarly, |−→F3y|=xsin60∘−ysin37∘ or, |−→F3y|=(√3x2−3y5) Since, −→F3 has no component in vertical axis, so, |−→F3y|=0 ⇒ (√3x2−3y5)=0 ⇒ √3x2=3y5⇒ x=2√3y5......(i) Also, |−−→F3x|=|−→F3| ⇒ (x2+4y5)=2 ⇒ 2√3y10+4y5=2 ⇒ y=10(√3+4) and, ⇒ x=2√3y5=4√3(√3+4) Hence, the magnitude of F1 and F2 respectively are 4√3(√3+4) N and 10(√3+4) N  Suggest Corrections  0    Join BYJU'S Learning Program
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