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
This answers would be 12x
8 x 5 = 40
40 x 18 = 720
Hope this helps!
Is that 2/3 of q or 2 over 3q?
for 2/3 of q. q=1/2 x 3/2 =3/4=0.75
for 2 over 3q q = 2 x 2/3= 4/3=1.3333
Multiply both sides by 2
(2) 18 = x/2 (2)
the /2 (2) gets canceled out
so it's 36= x
Hope this helps
