As AB is a diameter, then m∠CBA=180^{0}-75^{0}=105^{0}.
∠CBA is a central angle, so its measure multiplied by radius is the length of arc AC.
Note that
and
. Then
and length of the arc is
.The correct answer is 22 in.
Answer:
the answer is 441
Step-by-step explanation:
Part A
Represents 'Reflection'. This is so because the y-coordinates of P, Q and R remain the same in P' , Q' and R', and only the x-coordinate changes. Hence, it is reflection along the y-axis
Part B
Represents 'Rotation'. Here, the x-coordinates and y-coordinates of each of the points have changed, and the figure has been rotated clockwise around the point Q by 90°
Part C
Represents a combination of 'Translation' and 'Reflection'. Here either of the two has happened:
- First, all the points have been moved downwards by a fixed distance, thus changing the y-coordinate. Then, the resulting image has been reflected along the y-axis, thus changing the x-coordinate of all the points
- First, all the points have been moved to the right by a fixed distance, thus changing the x-coordinate. Then, the resulting image has been reflected along the x-axis, thus changing the y-coordinate of all the points
Part D
Represents 2 'Translations'. Here the image has been shifted by a fixed distance in both the downward direction and the right direction. Thus, it has resulted in change of both x and y coordinates.
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
