9514 1404 393
Answer:
- 2nd force: 99.91 lb
- resultant: 213.97 lb
Step-by-step explanation:
In the parallelogram shown, angle B is the supplement of angle DAB:
∠B = 180° -77°37' = 102°23'
Angle ACB is the difference of angles 77°37' and 27°8', so is 50°29'.
Now, we know the angles and one side of triangle ABC. We can use the law of sines to solve for the other two sides.
BC/sin(A) = AB/sin(C)
AD = BC = AB·sin(A)/sin(C) = (169 lb)sin(27°8')/sin(50°29') ≈ 99.91 lb
AC = AB·sin(B)/sin(C) = (169 lb)sin(102°23')/sin(50°29') ≈ 213.97 lb
Answer:
I think the answer is 60 or 70 or 65
Answer:
If every line parallel to two lines intersects both regions in line segments of equal length, then the two regions have equal areas. In the case of your problem, every line parallel to the bases of the two parallelograms will intersect them in lines segments, each with a width of ℓ.
