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Posted by **Hanson** on Tuesday, August 9, 2011 at 2:01pm.

The resultant of two forces acting on a body has a magnitude of 80 pounds. The angles between the resultant and the forces are 20° and 52°. Find the magnitude of the larger force.

- Precalculus -
**Damon**, Tuesday, August 9, 2011 at 2:36pma and b are forces

components in direction of resultant --> resultant

a cos 20 + b cos 52 = 80

components perpendicular to resultant sum to zero

a sin 20 = b sin 52

2 linear equations, two unknowns, solve

- Precalculus -
**Reiny**, Tuesday, August 9, 2011 at 2:40pmmake a sketch to get ∆ABC where ∢B = 20° and ∢A = 52°

AB is one vector, and AC is your second vector with BC the resultant.

so ∢C = 108°

By the sine law

AB/sin108 = 80/sin52

AB = 80sin108/sin52 = 96.55 or 97 pounds.

- Precalculus -
**Damon**, Tuesday, August 9, 2011 at 2:44pmQuestion #2

Which of us is a physicist and which is a mathematician ? :)

- Precalculus -
**Hanson**, Tuesday, August 9, 2011 at 2:49pmDamon, how would I solve thats what I am confused on

Reiny, That answer does not fit with any of the given answers

- Precalculus -
**Reiny**, Tuesday, August 9, 2011 at 3:00pmOk, let's do that triangle again.

draw AB the first vector, and AC, the second vector.

draw AP, the resultant.

angle BAP = 52

angle CAP = 20

AP = 80

complete the parallelogra BACP, then

angle APB=20

angle B = angle C = 108

AC/sin52 = 80/sin108

AC = 80sin52/sin108 = 66.3

(should not have jumped to a triangle too soon)

- Precalculus -
**Kate**, Tuesday, August 9, 2011 at 3:18pmThanks reiny! But one question how do you know when to use the triangle method and when to use parallelogram?

- Precalculus -
**Reiny**, Tuesday, August 9, 2011 at 3:36pmA good hint is to always draw the parallelogram first with the resultant being the diagonal.

That way you can use alternate angles (that is where I messed up)

The diagonal (resultant) will then give you the triangles.

Depending on the information given, you would then use either the sine law or the cosine law.