Asked by John

if two forces 10N and 20N are inclined at angle 60degree to each other. Find the resultant force.
Answered by Henry
F = 10N.[0o] + 20[60o] = 10 + 20*Cos60 +
i20*sin60 = 10 + 10 + 17.3i = 20 + 17.3i
= 26.5N.[40.9o].
Answered by Raymond
if two forces 10N and 20N are inclined at angle 60degree to each other. Find the resultant force.
Answered by BRIGHT
NOT CLEAR ENOUGH
Answered by Progress
R=f1+f2=10+20(both in square) R=100+400=500.(find the square root)=22.4n.therefore resultant force is 22.4n
Answered by Aminu Muhammed jamiu
Please
Answered by Happiness
R=20^2+10^ -2of10cos[120]=400+100-400[-0.5]=500+200=700
Answered by Anonymous
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Answered by Rebecca
I don't understand please explain better 😞
Answered by Bamise
Good
Answered by Sola
R²=400 +100 - 400cos120
=500+200
√700
R=26.5N
Answered by UME Chiemelie Sylvester
R^2= a^2+c^2-2ac cos(180 - THETA) Theta in this case is 60, 180-60=120
R^2 = 10^2 + 20^2 – (2 x 10 x 20 cos 120)
R^2 = 700N
R=√700N
R= √100 * √7 N
R= 10*√7 N or 10√7N

Answered by Habeeb
R=20^2+10^-2 of 10 cos [120]
=400+100-400[-0.5]
=500+200=700
Answered by Seema
√300
Answered by Francis
F1 is 10n while f2 is 20n while angle is 60°
r^2=20^2+10^2-2×20×10 cos 120
r^2=500+200
r^2=700
r^2=26.5
Answered by Zara♥️
R²=a²+b²-2ab cos theta
But R?
a=10N
b=20N
Theta=(180-60)=120°
R²=10²+20²-2×10×20 cos 120°
=100+400-400(0.5)
=500+400×0.5
=500+200
=700
R²=√700
=26.5N,,
Answered by Blessing
No. 1
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