Answer: find the solution in the explanation
Step-by-step explanation:
Let's use resolution of forces by resolving into x - component and y- component.
X - component.
Sum of forces = F1 - F3 - F4cos 15
Sum of forces = 0
5 - 5 - 0.97F4 = 0
- 0.97 F4 = 0
F4 = 0
Y - component
Sum of forces = F2 + F4 sin 15
Sum of forces = 0
5 + 0.26F4 = 0
0.26 F4 = -5
F4 = -5/0.26
F4 = -19.23 N
Simulating F4 back into the equation
Sum of forces = F1 - F3 - F4cos 15
- F4cos Ø = 0
- (-19.23) cos Ø = 0
Cos Ø = 0
Ø = 1
Does the angle formed approximate 15 degrees ? NO
Answer:
111°
Step-by-step explanation:
Let the centre of the circle be C
mRQ=157 (marked)
The angle at the centre of a circle standing on an arc is twice any angle at the circumference, standing on the same arc. So <SCR=2(SQR)=2(46)=92. mSR=<SCR=92
All the arc measure add up to 360 so:
mSQ+mRQ+mSR=360
mSQ+157+92=360
mSQ=360-249=111
Special Triangles Theorem
leg1=leg2=a
hypotenuse=a√2=10
Both Legs are 5√2 inches
It would be 1.276 x 10(5) the exponent is 5 next to the 10
By taking square root of a number using calculator