Three vectors

The three forces whose amplitudes are in ratio 9:10:17 act in the plane at one point so that they are in balance. Determine the angles of the each two forces.

Result

A =  53.13 °
B =  154.942 °
C =  151.928 °

Solution:

F1=9 F2=10 F3=17 u1=180πarccos((F22+F32F12)/(2 F2 F3))=180πarccos((102+17292)/(2 10 17))25.0576 u2=180πarccos((F12+F32F22)/(2 F1 F3))=180πarccos((92+172102)/(2 9 17))28.0725 A=u1+u2=25.0576+28.072553.130153.1353748"F_{1}=9 \ \\ F_{2}=10 \ \\ F_{3}=17 \ \\ u_{1}=\dfrac{ 180^\circ }{ \pi } \cdot \arccos((F_{2}^2+F_{3}^2-F_{1}^2)/(2 \cdot \ F_{2} \cdot \ F_{3}))=\dfrac{ 180^\circ }{ \pi } \cdot \arccos((10^2+17^2-9^2)/(2 \cdot \ 10 \cdot \ 17)) \doteq 25.0576 \ \\ u_{2}=\dfrac{ 180^\circ }{ \pi } \cdot \arccos((F_{1}^2+F_{3}^2-F_{2}^2)/(2 \cdot \ F_{1} \cdot \ F_{3}))=\dfrac{ 180^\circ }{ \pi } \cdot \arccos((9^2+17^2-10^2)/(2 \cdot \ 9 \cdot \ 17)) \doteq 28.0725 \ \\ A=u_{1}+u_{2}=25.0576+28.0725 \doteq 53.1301 \doteq 53.13 ^\circ \doteq 53^\circ 7'48"
B=180u1=18025.0576154.9424154.9421545633"B=180-u_{1}=180-25.0576 \doteq 154.9424 \doteq 154.942 ^\circ \doteq 154^\circ 56'33"
C=180u2=18028.0725151.9275151.9281515539"C=180-u_{2}=180-28.0725 \doteq 151.9275 \doteq 151.928 ^\circ \doteq 151^\circ 55'39"

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Tips to related online calculators
For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.
Check out our ratio calculator.
Two vectors given by its magnitudes and by included angle can be added by our vector sum calculator.
Cosine rule uses trigonometric SAS triangle calculator.
See also our trigonometric triangle calculator.

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