Solving Non-right Triangle
Vector Problems
(using right triangle trig)
Oh, Boy!
Here’s the problem:
Add the following displacements to get a
resultant displacement:
dA = 25.0m @ 33.0
dB = 42.0m @ 64.0
dres = ?
Step 1: Sketch the vectors and the resultant
dA + dB = dres
Add them head to
tail (start B from
the head of A)
Step 2: Resolve A & B into
perpendicular components
(x and y components)
By
Bx
Ay
Ax
Notice that if you add Bx to
Ax, they combine to make the
base of a big right triangle
whose hypotenuse is R!
You can do the
same with Ay
and By to make
the other side of
the big right
triangle.
By
Bx
Ay
Ax
Ay
Bx
Step 3: Solve for Ax and Bx
Ax = hyp (cosθ)
Ax = 25.0m(cos33.0 )
Ax = 21.0 m
By
Bx = hyp (cosθ)
Bx = 42.0m(cos64.0 )
Bx = 18.4 m
64.0
Bx
Ay
Ay
33.0
Ax= 21.0m
Bx= 18.4m
Step 4: Add Ax + Bx to make Rx—the
base of the big right triangle
Rx = Ax + Bx
Rx = 21.0m + 18.4m
Rx = 39.4 m
By
64.0
Bx
Ay
33.0
Ax
Bx
Rx = Ax + Bx = 39.4 m
Ay
Step 5: Solve for Ay and By
Ay = hyp (sinθ)
Ay = 25.0m(sin33.0 )
Ay = 13.6 m
By
= 37.7 m
By = hyp (sinθ)
By = 42.0m(sin64.0 )
By = 37.7 m
64.0
Bx
Ay
Ay
= 13.6 m
33.0
Ax
Bx
Ry = Ay + By
Ry = 13.6m + 37.7m
Ry = 51.3 m
By
64.0
Bx
Ay
Ay
33.0
Ax
Bx
Ry = Ay + By = 51.3 m
Step 6: Add Ay + By to make Ry—the
“height” of the big right triangle
c2 = a2 + b2
R2 = Rx2 + Ry2
R2 = (39.4m)2 + (51.3m)2
R = 64.5 m
Rx = Ax + Bx = 39.4 m
Ry = Ay + By = 51.3 m
Step 7: Solve for the hypotenuse
of the big right triangle—the
magnitude of the resultant
displacement
tan θ = opp./adj.
tan θ = 51.3m / 39.4m
θ = tan-1(51.3m / 39.4m)
θ = 52.5°
θ = 52.5°
Rx = Ax + Bx = 39.4 m
Ry = Ay + By = 51.3 m
Step 8: Solve for the direction of the
resultant vector by solving for angle
θ within the big triangle. Since you
have the lengths of all three sides
now, you can use sine, cosine or
tangent.
Final answer?
The resultant displacement
dres = 64.5 m @ 52.5°