Answer:
z = 110°
Step-by-step explanation:
Angles z and the one marked 70° form an linear pair, hence are supplementary.
z = 180° -70° = 110°
<h3>Angles where chords cross</h3>
The angle made by two chords is half the sum of the arcs intercepted by those chords. Here, that means ...
70° = 1/2(60° +x) ⇒ x = 140° -60° = 80°
The arc w completes the circle of 360°, so we have ...
w +x +79° +60° = 360° ⇒ w = 360° -219° = 141°
Finally, z is the average of w and 79°:
z = (w +79°)/2 = (141° +79°)/2 = 110° . . . as above
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B = C → A = C - B
→ B = C - A
Use the Double Angle Identity: cos 2A = 2 cos² A - 1
→ (cos 2A + 1)/2 = cos² A
Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]
Use Even/Odd Identity: cos (-A) = cos (A)
<u>Proof LHS → RHS:</u>
LHS: cos² A + cos² B + cos² C
![\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cquad%201%2B%5Cdfrac%7B1%7D%7B2%7D%5Cbigg%5B2%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A%2B2B%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A-2B%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%2B%5Ccos%5E2%20C%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D1%2B%5Ccos%20%28A%2BB%29%5Ccdot%20%5Ccos%20%28A-B%29%2B%5Ccos%5E2%20C)
![\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%20%5Cqquad%201%2B%5Ccos%20C%5B%5Ccos%20%28A-B%29%2B%5Ccos%20C%5D)
![\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5Cquad%201%2B%5Ccos%20C%5Cbigg%5B2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%2BC%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B-C%7D%7B2%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D1%2B2%5Ccos%20C%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2B%28C-B%29%7D%7B2%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B-B-%28C-A%29%7D%7B2%7D%5Cbigg%29)
LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C 
M<1+ m<2 = 90
-2x + 54 + 8x + 18 = 90
6x + 72 = 90
6x = 90 - 72
6x = 18
x = 3
Answer:
The remaining is Five ?
Step-by-step explanation:
Answer:
the answer for me would be (p+q)+r
here is why I think so
q +pr ( the pr close-by together in maths means multiple so that's out) same goes for C and A it has a negative so that's not equal.
