$A_x = 40N\;, A_y = ?,\; q = 60^°$

$\therefore {A_x}\,\, = \,\,A\,\,\cos \,\theta \,\,\,\,\therefore \,\,40\,\, = \,A\,\,\cos \,\,60^\circ \,\, = \,\,\,\frac{A}{2}\,\,$ અથવા $\,A\,\, = \,\,80N$

${A_y}\,\, = \,\,A\,\,\sin \,\,60^\circ \,\, = \,\,\frac{{A\sqrt 3 }}{2}\,\, = \,\,\frac{{80\sqrt 3 }}{2}\,\, = \,\,40\sqrt 3 \,\,N\,\, = \,\,40\,\, \times \,\,1.732\,\, = \,\,69.28\,\,N$